Tr\u01b0\u1eddng THPT Ho\u00e0ng L\u1ec7 Kha, Thanh H\u00f3a t\u1ed5 ch\u1ee9c thi th\u1eed THPT Qu\u1ed1c gia n\u0103m 2016 m\u00f4n L\u00fd m\u00e3 \u0111\u1ec1 132, c\u00f3 \u0111\u00e1p \u00e1n, xem chi ti\u1ebft \u0111\u1ec1 thi v\u00e0 \u0111\u00e1p \u00e1n d\u01b0\u1edbi \u0111\u00e2y:<\/strong><\/h2>\n\u0110\u1ec1 thi th\u1eed THPTQG m\u00f4n L\u00fd THPT Ho\u00e0ng L\u1ec7 Kha 2016<\/strong><\/h3>\n![](\"http:\/\/img.toanhoc247.com\/picture\/2015\/1128\/de-thi-thu-thptqg-ly-hoang-le-kha-5.png\")
<\/strong><\/p>\nC\u00e2u 14:<\/strong>S\u00f3ng \u0111i\u1ec7n t\u1eeb<\/p>\nA.<\/strong>l\u00e0 s\u00f3ng d\u1ecdc ho\u1eb7c s\u00f3ng ngang.<\/p>\nB.<\/strong>l\u00e0 \u0111i\u1ec7n t\u1eeb tr\u01b0\u1eddng lan truy\u1ec1n trong kh\u00f4ng gian.<\/p>\nC.<\/strong>c\u00f3 th\u00e0nh ph\u1ea7n \u0111i\u1ec7n tr\u01b0\u1eddng v\u00e0 th\u00e0nh ph\u1ea7n t\u1eeb tr\u01b0\u1eddng t\u1ea1i m\u1ed9t \u0111i\u1ec3m dao \u0111\u1ed9ng c\u00f9ng ph\u01b0\u01a1ng.<\/p>\nD.<\/strong>kh\u00f4ng truy\u1ec1n \u0111\u01b0\u1ee3c trong ch\u00e2n kh\u00f4ng<\/p>\nC\u00e2u 15:<\/strong>Cho 2 dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u00f9ng ph\u01b0\u01a1ng, c\u00f9ng t\u1ea7n s\u1ed1 c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x<\/p>\n1<\/dd>\n= A<\/p>\n
1<\/dd>\ncos(wt +j<\/p>\n
1<\/dd>\n); x<\/p>\n
2<\/dd>\n= A<\/p>\n
2<\/dd>\ncos(wt + j<\/p>\n
2<\/dd>\n). Bi\u00ean \u0111\u1ed9 dao \u0111\u1ed9ng t\u1ed5ng h\u1ee3p c\u00f3 gi\u00e1 c\u1ef1c \u0111\u1ea1i khi<\/p>\n
A.<\/strong>Hai dao \u0111\u1ed9ng c\u00f9ng pha B.<\/strong>Hai dao \u0111\u1ed9ng ng\u01b0\u1ee3c pha<\/p>\nC.<\/strong>Hai dao \u0111\u1ed9ng vu\u00f4ng pha D.<\/strong>Hai dao \u0111\u1ed9ng \u2113\u1ec7ch pha 120<\/p>\n0<\/dd>\n<\/p>\nC\u00e2u 16:<\/strong>M\u1ed9t h\u1ecdc sinh \u0111o gia t\u1ed1c tr\u1ecdng tr\u01b0\u1eddng t\u1ea1i v\u1ecb tr\u00ed \u0111\u1ecba l\u00ed n\u01a1i tr\u01b0\u1eddng \u0111\u1eb7t \u0111\u1ecba \u0111i\u1ec3m th\u00f4ng qua vi\u1ec7c \u0111o chu k\u00ec dao \u0111\u1ed9ng c\u1ee7a con l\u1eafc l\u00f2 xo g\u1ed3m m\u1ed9t l\u00f2 xo c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng kh\u00f4ng \u0111\u00e1ng k\u1ec3 \u0111\u1ea7u tr\u00ean c\u1ed1 \u0111\u1ecbnh, \u0111\u1ea7u d\u01b0\u1edbi g\u1eafn m\u1ed9t qu\u1ea3 c\u1ea7u nh\u1ecf. K\u00edch th\u00edch cho con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng theo ph\u01b0\u01a1ng th\u1eb3ng \u0111\u1ee9ng v\u00e0 d\u00f9ng \u0111\u1ed3ng h\u1ed3 b\u1ea5m d\u00e2y h\u1ecdc sinh \u0111o \u0111\u01b0\u1ee3c chu k\u00ec dao \u0111\u1ed9ng c\u1ee7a qu\u1ea3 c\u1ea7u l\u00e0 T = ( 0,69 0,01 ) s. D\u00f9ng th\u01b0\u1edbc h\u1ecdc sinh n\u00e0y \u0111o \u0111\u01b0\u1ee3c \u0111\u1ed9 d\u00e3n c\u1ee7a l\u00f2 xo khi qu\u1ea3 c\u1ea7u \u0111\u1ee9ng c\u00e2n b\u1eb1ng l\u00e0 x = ( 119,5 0,5 ) mm. L\u1ea5y = 3,14. Sai s\u1ed1 t\u1ec9 \u0111\u1ed1i c\u1ee7a ph\u00e9p \u0111o gia t\u1ed1c tr\u1ecdng tr\u01b0\u1eddng l\u00e0<\/p>\nA.<\/strong>3,31%. B.<\/strong>1,87%. C.<\/strong>1,03%. D.<\/strong>2,48%.<\/p>\nC\u00e2u 17:<\/strong>M\u1ed9t con l\u1eafc l\u00f2 xo n\u1eb1m ngang dao \u0111\u1ed9ng t\u1ef1 do v\u1edbi bi\u00ean \u0111\u1ed9 8 cm. L\u1ef1c \u0111\u00e0n h\u1ed3i c\u1ee7a l\u00f2 xo c\u00f3 c\u00f4ng su\u1ea5t t\u1ee9c th\u1eddi \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i khi v\u1eadt \u0111i qua v\u1ecb tr\u00ed c\u00f3 to\u1ea1 \u0111\u1ed9 x b\u1eb1ng.<\/p>\nA.<\/strong>0 B.<\/strong> 4 cm C.<\/strong> 8 cm D.<\/strong> 4cm<\/p>\nC\u00e2u 18:<\/strong>Con l\u1eafc l\u00f2 xo treo th\u1eb3ng \u0111\u1ee9ng. \u0110i\u1ec3m treo l\u00e0 O. \u0110\u1ed9 c\u1ee9ng l\u00f2 xo l\u00e0 10N\/m. T\u1eeb v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng, n\u00e2ng v\u1eadt l\u00ean \u0111o\u1ea1n 30cm r\u1ed3i bu\u00f4ng nh\u1eb9 cho v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a th\u00ec th\u1ea5y chu k\u1ef3 dao \u0111\u1ed9ng c\u1ee7a v\u1eadt l\u00e0 1 gi\u00e2y. L\u1ea5y g = 10(m\/s<\/p>\n2<\/dd>\n) = \u03c0<\/p>\n
2<\/dd>\n(m\/s<\/p>\n
2<\/dd>\n). L\u1ef1c \u0111\u1ea9y c\u1ef1c \u0111\u1ea1i t\u00e1c d\u1ee5ng l\u00ean \u0111i\u1ec3m O l\u00e0:<\/p>\n
A.<\/strong>1,25 N B.<\/strong>1,55 N C.<\/strong>0,5 N D.<\/strong>0,55 N<\/p>\nC\u00e2u 19:<\/strong>M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a d\u1ecdc theo tr\u1ee5c Ox v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh x = 5cos(8 pt + p\/2)(cm). Chi\u1ec1u d\u00e0i qu\u1ef9 \u0111\u1ea1o c\u1ee7a v\u1eadt l\u00e0<\/p>\nA.<\/strong>5cm. B.<\/strong>20cm. C.<\/strong>2,5cm. D.<\/strong>10cm.<\/p>\nC\u00e2u 20:<\/strong>Hai v\u1eadt A v\u00e0 B c\u00f3 c\u00f9ng kh\u1ed1i l\u01b0\u1ee3ng 1 kg v\u00e0 c\u00f3 k\u00edch th\u01b0\u1edbc nh\u1ecf \u0111\u01b0\u1ee3c n\u1ed1i v\u1edbi nhau b\u1edfi s\u1ee3i d\u00e2y m\u1ea3nh nh\u1eb9 d\u00e0i 10cm, hai v\u1eadt \u0111\u01b0\u1ee3c treo v\u00e0o l\u00f2 xo c\u00f3 \u0111\u1ed9 c\u1ee9ng k = 100 N\/m t\u1ea1i n\u01a1i c\u00f3 gia t\u1ed1c tr\u1ecdng tr\u01b0\u1eddng g = 10 m\/s<\/p>\n2<\/dd>\n. L\u1ea5y \u03c0<\/p>\n
2<\/dd>\n= 10. Khi h\u1ec7 v\u1eadt v\u00e0 l\u00f2 xo \u0111ang \u1edf v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng ng\u01b0\u1eddi ta \u0111\u1ed1t s\u1ee3i d\u00e2y n\u1ed1i hai v\u1eadt v\u00e0 v\u1eadt B s\u1ebd r\u01a1i t\u1ef1 do c\u00f2n v\u1eadt A s\u1ebd dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a. L\u1ea7n \u0111\u1ea7u ti\u00ean v\u1eadt A l\u00ean \u0111\u1ebfn v\u1ecb tr\u00ed cao nh\u1ea5t th\u00ec kho\u1ea3ng c\u00e1ch gi\u1eefa hai v\u1eadt b\u1eb1ng bao nhi\u00eau? Bi\u1ebft r\u1eb1ng \u0111\u1ed9 cao \u0111\u1ee7 l\u1edbn.<\/p>\n
A.<\/strong>20 cm. B.<\/strong>50 cm. C.<\/strong>80 cm. D.<\/strong>70 cm.<\/p>\nC\u00e2u 21:<\/strong>S\u00f3ng truy\u1ec1n v\u1edbi t\u1ed1c \u0111\u1ed9 kh\u00f4ng \u0111\u1ed5i 10m\/s t\u1eeb \u0111i\u1ec3m M \u0111\u1ebfn O tr\u00ean c\u00f9ng ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng v\u1edbi MO = 50cm, coi bi\u00ean \u0111\u1ed9 s\u00f3ng kh\u00f4ng \u0111\u1ed5i. Bi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh s\u00f3ng t\u1ea1i O l\u00e0 u<\/p>\nO<\/dd>\n= 5cos(10\u03c0t) cm. Ph\u01b0\u01a1ng tr\u00ecnh s\u00f3ng t\u1ea1i M l\u00e0:<\/p>\n
A.<\/strong>u = 5cos(10\u03c0t – \u03c0\/4) cm B.<\/strong>u = 5cos(10\u03c0t – \u03c0\/2) cm<\/p>\nC.<\/strong>u = 5cos(10\u03c0t + \u03c0\/6) cm D.<\/strong>u = 5cos(10\u03c0t + \u03c0\/2) cm<\/p>\nC\u00e2u 22:<\/strong>B\u01b0\u1edbc s\u00f3ng l\u00e0 kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111i\u1ec3m<\/p>\nA.<\/strong>g\u1ea7n nhau nh\u1ea5t m\u00e0 dao \u0111\u1ed9ng t\u1ea1i hai \u0111i\u1ec3m \u0111\u00f3 c\u00f9ng pha.<\/p>\nB.<\/strong>g\u1ea7n nhau nh\u1ea5t tr\u00ean c\u00f9ng m\u1ed9t ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng m\u00e0 dao \u0111\u1ed9ng t\u1ea1i hai \u0111i\u1ec3m \u0111\u00f3 c\u00f9ng pha.<\/p>\nC.<\/strong>tr\u00ean c\u00f9ng m\u1ed9t ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng m\u00e0 dao \u0111\u1ed9ng t\u1ea1i hai \u0111i\u1ec3m \u0111\u00f3 ng\u01b0\u1ee3c pha.<\/p>\nD.<\/strong>tr\u00ean c\u00f9ng m\u1ed9t ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng m\u00e0 dao \u0111\u1ed9ng t\u1ea1i hai \u0111i\u1ec3m \u0111\u00f3 c\u00f9ng pha.<\/p>\nC\u00e2u 23:<\/strong>M\u1ed9t s\u1ee3i d\u00e2y \u0111\u00e0n h\u1ed3i AB d\u00e0i 2m \u0111\u01b0\u1ee3c c\u0103ng ngang, \u0111\u1ea7u B c\u1ed1 \u0111\u1ecbnh, \u0111\u1ea7u A g\u1eafn v\u00e0o 1 c\u1ea7n rung dao \u0111\u1ed9ng v\u1edbi t\u1ea7n s\u1ed1 50Hz. T\u1ed1c \u0111\u1ed9 truy\u1ec1n s\u00f3ng tr\u00ean d\u00e2y l\u00e0 50m\/s. Tr\u00ean d\u00e2y h\u00ecnh th\u00e0nh s\u00f3ng d\u1eebng v\u1edbi<\/p>\nA.<\/strong>3 b\u1ee5ng, 4 n\u00fat. B.<\/strong>2 b\u1ee5ng, 3 n\u00fat. C.<\/strong>4 b\u1ee5ng, 5 n\u00fat. D.<\/strong>1 b\u1ee5ng, 2 n\u00fat.<\/p>\nC\u00e2u 24:<\/strong>Con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a v\u1edbi t\u1ea7n s\u1ed1 f. \u0110\u1ed9ng n\u0103ng v\u00e0 th\u1ebf n\u0103ng c\u1ee7a con l\u1eafc bi\u1ebfn thi\u00ean tu\u1ea7n ho\u00e0n v\u1edbi t\u1ea7n s\u1ed1 l\u00e0<\/p>\nA.<\/strong>2f. B.<\/strong>f. C.<\/strong>4f. D.<\/strong>f\/2.<\/p>\nC\u00e2u 25:<\/strong>Hai \u00e2m c\u00f9ng \u0111\u1ed9 cao l\u00e0 hai \u00e2m c\u00f3 c\u00f9ng<\/p>\nA.<\/strong>bi\u00ean \u0111\u1ed9. B.<\/strong>t\u1ea7n s\u1ed1. C.<\/strong>c\u01b0\u1eddng \u0111\u1ed9 \u00e2m. D.<\/strong>m\u1ee9c c\u01b0\u1eddng \u0111\u1ed9 \u00e2m.<\/p>\nC\u00e2u 26:<\/strong>Nh\u1eadn \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0sai<\/u><\/strong>khi n\u00f3i v\u1ec1 dao \u0111\u1ed9ng c\u01a1 h\u1ecdc t\u1eaft d\u1ea7n?<\/p>\nA.<\/strong>Dao \u0111\u1ed9ng t\u1eaft d\u1ea7n l\u00e0 dao \u0111\u1ed9ng c\u00f3 bi\u00ean \u0111\u1ed9 gi\u1ea3m d\u1ea7n theo th\u1eddi gian.<\/p>\nB.<\/strong>Trong dao \u0111\u1ed9ng t\u1eaft d\u1ea7n, c\u01a1 n\u0103ng gi\u1ea3m d\u1ea7n theo th\u1eddi gian.<\/p>\nC.<\/strong>L\u1ef1c ma s\u00e1t c\u00e0ng l\u1edbn th\u00ec dao \u0111\u1ed9ng t\u1eaft c\u00e0ng nhanh.<\/p>\nD.<\/strong>Dao<\/u>\u0111\u1ed9ng t\u1eaft d\u1ea7n c\u00f3 \u0111\u1ed9ng n\u0103ng gi\u1ea3m d\u1ea7n c\u00f2n th\u1ebf n\u0103ng bi\u1ebfn thi\u00ean \u0111i\u1ec1u h\u00f2a<\/p>\nC\u00e2u 27:<\/strong>M\u1ed9t con l\u1eafc \u0111\u01a1n d\u00e0i l = 25cm, h\u00f2n bi c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng m = 10g v\u00e0 mang \u0111i\u1ec7n t\u00edch q = 10<\/p>\n-4<\/dd>\nC. Treo con l\u1eafc v\u00e0o gi\u1eefa hai b\u1ea3n kim lo\u1ea1i th\u1eb3ng \u0111\u1ee9ng, song song c\u00e1ch nhau d = 22cm. \u0110\u1eb7t v\u00e0o hai b\u1ea3n hi\u1ec7u \u0111i\u1ec7n th\u1ebf m\u1ed9t chi\u1ec1u U = 88V, l\u1ea5y g = 10 m\/s<\/p>\n
2<\/dd>\n. Chu k\u00ec dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a v\u1edbi bi\u00ean \u0111\u1ed9 nh\u1ecf c\u1ee7a con l\u1eafc \u0111\u01a1n l\u00e0:<\/p>\n
A.<\/strong>0,659s B.<\/strong>0,983 s C.<\/strong>0,897s D.<\/strong>0,957 s<\/p>\nC\u00e2u 28:<\/strong>Hai ngu\u1ed3n s\u00f3ng c\u01a1 k\u1ebft h\u1ee3p S<\/p>\n1<\/dd>\n, S<\/p>\n
2<\/dd>\ndao \u0111\u1ed9ng c\u00f9ng ph\u01b0\u01a1ng, c\u00f9ng bi\u00ean \u0111\u1ed9 v\u00e0 ng\u01b0\u1ee3c pha c\u00e1ch nhau 60 cm c\u00f3 t\u1ea7n s\u1ed1 5 Hz, t\u1ed1c \u0111\u1ed9 truy\u1ec1n s\u00f3ng l\u00e0 40 cm\/s. S\u1ed1 c\u1ef1c ti\u1ec3u giao thoa tr\u00ean \u0111o\u1ea1n S<\/p>\n
1<\/dd>\nS<\/p>\n
2<\/dd>\nl\u00e0:<\/p>\n
A.<\/strong>15. B.<\/strong>16. C.<\/strong>14. D.<\/strong>13.<\/p>\nC\u00e2u 29:<\/strong>Con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0, khi t\u0103ng kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a v\u1eadt l\u00ean 4 l\u1ea7n th\u00ec chu k\u00ec dao \u0111\u1ed9ng c\u1ee7a v\u1eadt<\/p>\nA.<\/strong>t\u0103ng l\u00ean 2 l\u1ea7n. B.<\/strong>gi\u1ea3m \u0111i 2 l\u1ea7n. C.<\/strong>gi\u1ea3m \u0111i 4 l\u1ea7n. D.<\/strong>t\u0103ng l\u00ean 4 l\u1ea7n.<\/p>\nC\u00e2u 30:<\/strong>Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng c\u1ee7a m\u1ed9t v\u1eadt c\u00f3 d\u1ea1ng x = Asin(t +\/4). Ch\u1ecdn k\u1ebft lu\u1eadn\u0111\u00fang<\/em>.<\/p>\nA.<\/strong>V\u1eadt dao \u0111\u1ed9ng v\u1edbi bi\u00ean \u0111\u1ed9 A. B.<\/strong>V\u1eadt dao \u0111\u1ed9ng v\u1edbi pha ban \u0111\u1ea7u\/4.<\/p>\nC.<\/strong>V\u1eadt dao \u0111\u1ed9ng v\u1edbi bi\u00ean \u0111\u1ed9 2A. D.<\/strong>V\u1eadt dao \u0111\u1ed9ng v\u1edbi bi\u00ean \u0111\u1ed9 A\/2.<\/p>\nC\u00e2u 31:<\/strong>Ngu\u1ed3n \u00e2m S ph\u00e1t ra \u00e2m c\u00f3 c\u00f4ng su\u1ea5t kh\u00f4ng \u0111\u1ed5i truy\u1ec1n \u0111\u1eb3ng h\u01b0\u1edbng v\u1ec1 m\u1ecdi ph\u01b0\u01a1ng. T\u1ea1i M c\u00e1ch S \u0111o\u1ea1n 2m, m\u1ee9c c\u01b0\u1eddng \u0111\u1ed9 \u00e2m l\u00e0 50 dB. Gi\u1ea3 s\u1eed m\u00f4i tr\u01b0\u1eddng kh\u00f4ng h\u1ea5p th\u1ee5 \u00e2m. M\u1ee9c c\u01b0\u1eddng \u0111\u1ed9 \u00e2m t\u1ea1i N c\u00e1ch S \u0111o\u1ea1n 8m l\u00e0:<\/p>\nA.<\/strong>38 dB. B.<\/strong>48 dB. C.<\/strong>42 dB. D.<\/strong>45 dB.<\/p>\n![](\"http:\/\/img.toanhoc247.com\/picture\/2015\/1128\/de-thi-thu-thptqg-ly-hoang-le-kha-6.png\")
<\/p>\n
\u0110\u00e1p \u00e1n \u0111\u1ec1 thi th\u1eed THPTQG m\u00f4n L\u00fd THPT Ho\u00e0ng L\u1ec7 Kha 2016<\/strong><\/h3>\n\n\n\n1<\/td>\n C<\/td>\n 16<\/td>\n A<\/td>\n 31<\/td>\n A<\/td>\n 46<\/td>\n B<\/td>\n<\/tr>\n \n2<\/td>\n D<\/td>\n 17<\/td>\n D<\/td>\n 32<\/td>\n B<\/td>\n 47<\/td>\n B<\/td>\n<\/tr>\n \n3<\/td>\n A<\/td>\n 18<\/td>\n C<\/td>\n 33<\/td>\n C<\/td>\n 48<\/td>\n D<\/td>\n<\/tr>\n \n4<\/td>\n C<\/td>\n 19<\/td>\n D<\/td>\n 34<\/td>\n C<\/td>\n 49<\/td>\n D<\/td>\n<\/tr>\n \n5<\/td>\n B<\/td>\n 20<\/td>\n C<\/td>\n 35<\/td>\n B<\/td>\n 50<\/td>\n C<\/td>\n<\/tr>\n \n6<\/td>\n B<\/td>\n 21<\/td>\n D<\/td>\n 36<\/td>\n D<\/td>\n<\/tr>\n \n7<\/td>\n B<\/td>\n 22<\/td>\n B<\/td>\n 37<\/td>\n D<\/td>\n<\/tr>\n \n8<\/td>\n C<\/td>\n 23<\/td>\n C<\/td>\n 38<\/td>\n C<\/td>\n<\/tr>\n \n9<\/td>\n B<\/td>\n 24<\/td>\n A<\/td>\n 39<\/td>\n A<\/td>\n<\/tr>\n \n10<\/td>\n B<\/td>\n 25<\/td>\n B<\/td>\n 40<\/td>\n B<\/td>\n<\/tr>\n \n11<\/td>\n D<\/td>\n 26<\/td>\n D<\/td>\n 41<\/td>\n A<\/td>\n<\/tr>\n \n12<\/td>\n D<\/td>\n 27<\/td>\n D<\/td>\n 42<\/td>\n D<\/td>\n<\/tr>\n \n13<\/td>\n A<\/td>\n 28<\/td>\n A<\/td>\n 43<\/td>\n C<\/td>\n<\/tr>\n \n14<\/td>\n B<\/td>\n 29<\/td>\n A<\/td>\n 44<\/td>\n C<\/td>\n<\/tr>\n \n15<\/td>\n A<\/td>\n 30<\/td>\n A<\/td>\n 45<\/td>\n A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
<\/strong><\/p>\nT\u1ea5t c\u1ea3 n\u1ed9i dung b\u00e0i vi\u1ebft. C\u00e1c em h\u00e3y xem th\u00eam v\u00e0 t\u1ea3i file chi ti\u1ebft t\u1ea1i \u0111\u00e2y:<\/strong>Download<\/p>\n","protected":false},"excerpt":{"rendered":"Tr\u01b0\u1eddng THPT Ho\u00e0ng L\u1ec7 Kha, Thanh H\u00f3a t\u1ed5 ch\u1ee9c thi th\u1eed THPT Qu\u1ed1c gia n\u0103m 2016 m\u00f4n L\u00fd m\u00e3 \u0111\u1ec1 132, c\u00f3 \u0111\u00e1p \u00e1n, xem chi ti\u1ebft \u0111\u1ec1 thi v\u00e0 \u0111\u00e1p \u00e1n d\u01b0\u1edbi \u0111\u00e2y: \u0110\u1ec1 thi th\u1eed THPTQG m\u00f4n L\u00fd THPT Ho\u00e0ng L\u1ec7 Kha 2016 C\u00e2u 14:S\u00f3ng \u0111i\u1ec7n t\u1eeb A.l\u00e0 s\u00f3ng d\u1ecdc ho\u1eb7c s\u00f3ng ngang. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[25],"tags":[],"yoast_head":"\n
\u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n L\u00fd THPT Ho\u00e0ng L\u1ec7 Kha 2016 c\u00f3 \u0111\u00e1p \u00e1n<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/de-thi-thu-thpt-quoc-gia-mon-ly-thpt-hoang-le-kha-2016-co-dap-an\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"\u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n L\u00fd THPT Ho\u00e0ng L\u1ec7 Kha 2016 c\u00f3 \u0111\u00e1p \u00e1n\" \/>\n<meta property=\"og:description\" content=\"Tr\u01b0\u1eddng THPT Ho\u00e0ng L\u1ec7 Kha, Thanh H\u00f3a t\u1ed5 ch\u1ee9c thi th\u1eed THPT Qu\u1ed1c gia n\u0103m 2016 m\u00f4n L\u00fd m\u00e3 \u0111\u1ec1 132, c\u00f3 \u0111\u00e1p \u00e1n, xem chi ti\u1ebft \u0111\u1ec1 thi v\u00e0 \u0111\u00e1p \u00e1n d\u01b0\u1edbi \u0111\u00e2y: \u0110\u1ec1 thi th\u1eed THPTQG m\u00f4n L\u00fd THPT Ho\u00e0ng L\u1ec7 Kha 2016 C\u00e2u 14:S\u00f3ng \u0111i\u1ec7n t\u1eeb A.l\u00e0 s\u00f3ng d\u1ecdc ho\u1eb7c s\u00f3ng ngang. 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C\u00e2u 14:<\/strong>S\u00f3ng \u0111i\u1ec7n t\u1eeb<\/p>\n A.<\/strong>l\u00e0 s\u00f3ng d\u1ecdc ho\u1eb7c s\u00f3ng ngang.<\/p>\n B.<\/strong>l\u00e0 \u0111i\u1ec7n t\u1eeb tr\u01b0\u1eddng lan truy\u1ec1n trong kh\u00f4ng gian.<\/p>\n C.<\/strong>c\u00f3 th\u00e0nh ph\u1ea7n \u0111i\u1ec7n tr\u01b0\u1eddng v\u00e0 th\u00e0nh ph\u1ea7n t\u1eeb tr\u01b0\u1eddng t\u1ea1i m\u1ed9t \u0111i\u1ec3m dao \u0111\u1ed9ng c\u00f9ng ph\u01b0\u01a1ng.<\/p>\n D.<\/strong>kh\u00f4ng truy\u1ec1n \u0111\u01b0\u1ee3c trong ch\u00e2n kh\u00f4ng<\/p>\n C\u00e2u 15:<\/strong>Cho 2 dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u00f9ng ph\u01b0\u01a1ng, c\u00f9ng t\u1ea7n s\u1ed1 c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x<\/p>\n = A<\/p>\n cos(wt +j<\/p>\n ); x<\/p>\n = A<\/p>\n cos(wt + j<\/p>\n ). Bi\u00ean \u0111\u1ed9 dao \u0111\u1ed9ng t\u1ed5ng h\u1ee3p c\u00f3 gi\u00e1 c\u1ef1c \u0111\u1ea1i khi<\/p>\n A.<\/strong>Hai dao \u0111\u1ed9ng c\u00f9ng pha B.<\/strong>Hai dao \u0111\u1ed9ng ng\u01b0\u1ee3c pha<\/p>\n C.<\/strong>Hai dao \u0111\u1ed9ng vu\u00f4ng pha D.<\/strong>Hai dao \u0111\u1ed9ng \u2113\u1ec7ch pha 120<\/p>\n C\u00e2u 16:<\/strong>M\u1ed9t h\u1ecdc sinh \u0111o gia t\u1ed1c tr\u1ecdng tr\u01b0\u1eddng t\u1ea1i v\u1ecb tr\u00ed \u0111\u1ecba l\u00ed n\u01a1i tr\u01b0\u1eddng \u0111\u1eb7t \u0111\u1ecba \u0111i\u1ec3m th\u00f4ng qua vi\u1ec7c \u0111o chu k\u00ec dao \u0111\u1ed9ng c\u1ee7a con l\u1eafc l\u00f2 xo g\u1ed3m m\u1ed9t l\u00f2 xo c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng kh\u00f4ng \u0111\u00e1ng k\u1ec3 \u0111\u1ea7u tr\u00ean c\u1ed1 \u0111\u1ecbnh, \u0111\u1ea7u d\u01b0\u1edbi g\u1eafn m\u1ed9t qu\u1ea3 c\u1ea7u nh\u1ecf. K\u00edch th\u00edch cho con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng theo ph\u01b0\u01a1ng th\u1eb3ng \u0111\u1ee9ng v\u00e0 d\u00f9ng \u0111\u1ed3ng h\u1ed3 b\u1ea5m d\u00e2y h\u1ecdc sinh \u0111o \u0111\u01b0\u1ee3c chu k\u00ec dao \u0111\u1ed9ng c\u1ee7a qu\u1ea3 c\u1ea7u l\u00e0 T = ( 0,69 0,01 ) s. D\u00f9ng th\u01b0\u1edbc h\u1ecdc sinh n\u00e0y \u0111o \u0111\u01b0\u1ee3c \u0111\u1ed9 d\u00e3n c\u1ee7a l\u00f2 xo khi qu\u1ea3 c\u1ea7u \u0111\u1ee9ng c\u00e2n b\u1eb1ng l\u00e0 x = ( 119,5 0,5 ) mm. L\u1ea5y = 3,14. Sai s\u1ed1 t\u1ec9 \u0111\u1ed1i c\u1ee7a ph\u00e9p \u0111o gia t\u1ed1c tr\u1ecdng tr\u01b0\u1eddng l\u00e0<\/p>\n A.<\/strong>3,31%. B.<\/strong>1,87%. C.<\/strong>1,03%. D.<\/strong>2,48%.<\/p>\n C\u00e2u 17:<\/strong>M\u1ed9t con l\u1eafc l\u00f2 xo n\u1eb1m ngang dao \u0111\u1ed9ng t\u1ef1 do v\u1edbi bi\u00ean \u0111\u1ed9 8 cm. L\u1ef1c \u0111\u00e0n h\u1ed3i c\u1ee7a l\u00f2 xo c\u00f3 c\u00f4ng su\u1ea5t t\u1ee9c th\u1eddi \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i khi v\u1eadt \u0111i qua v\u1ecb tr\u00ed c\u00f3 to\u1ea1 \u0111\u1ed9 x b\u1eb1ng.<\/p>\n A.<\/strong>0 B.<\/strong> 4 cm C.<\/strong> 8 cm D.<\/strong> 4cm<\/p>\n C\u00e2u 18:<\/strong>Con l\u1eafc l\u00f2 xo treo th\u1eb3ng \u0111\u1ee9ng. \u0110i\u1ec3m treo l\u00e0 O. \u0110\u1ed9 c\u1ee9ng l\u00f2 xo l\u00e0 10N\/m. T\u1eeb v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng, n\u00e2ng v\u1eadt l\u00ean \u0111o\u1ea1n 30cm r\u1ed3i bu\u00f4ng nh\u1eb9 cho v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a th\u00ec th\u1ea5y chu k\u1ef3 dao \u0111\u1ed9ng c\u1ee7a v\u1eadt l\u00e0 1 gi\u00e2y. L\u1ea5y g = 10(m\/s<\/p>\n ) = \u03c0<\/p>\n (m\/s<\/p>\n ). L\u1ef1c \u0111\u1ea9y c\u1ef1c \u0111\u1ea1i t\u00e1c d\u1ee5ng l\u00ean \u0111i\u1ec3m O l\u00e0:<\/p>\n A.<\/strong>1,25 N B.<\/strong>1,55 N C.<\/strong>0,5 N D.<\/strong>0,55 N<\/p>\n C\u00e2u 19:<\/strong>M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a d\u1ecdc theo tr\u1ee5c Ox v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh x = 5cos(8 pt + p\/2)(cm). Chi\u1ec1u d\u00e0i qu\u1ef9 \u0111\u1ea1o c\u1ee7a v\u1eadt l\u00e0<\/p>\n A.<\/strong>5cm. B.<\/strong>20cm. C.<\/strong>2,5cm. D.<\/strong>10cm.<\/p>\n C\u00e2u 20:<\/strong>Hai v\u1eadt A v\u00e0 B c\u00f3 c\u00f9ng kh\u1ed1i l\u01b0\u1ee3ng 1 kg v\u00e0 c\u00f3 k\u00edch th\u01b0\u1edbc nh\u1ecf \u0111\u01b0\u1ee3c n\u1ed1i v\u1edbi nhau b\u1edfi s\u1ee3i d\u00e2y m\u1ea3nh nh\u1eb9 d\u00e0i 10cm, hai v\u1eadt \u0111\u01b0\u1ee3c treo v\u00e0o l\u00f2 xo c\u00f3 \u0111\u1ed9 c\u1ee9ng k = 100 N\/m t\u1ea1i n\u01a1i c\u00f3 gia t\u1ed1c tr\u1ecdng tr\u01b0\u1eddng g = 10 m\/s<\/p>\n . L\u1ea5y \u03c0<\/p>\n = 10. Khi h\u1ec7 v\u1eadt v\u00e0 l\u00f2 xo \u0111ang \u1edf v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng ng\u01b0\u1eddi ta \u0111\u1ed1t s\u1ee3i d\u00e2y n\u1ed1i hai v\u1eadt v\u00e0 v\u1eadt B s\u1ebd r\u01a1i t\u1ef1 do c\u00f2n v\u1eadt A s\u1ebd dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a. L\u1ea7n \u0111\u1ea7u ti\u00ean v\u1eadt A l\u00ean \u0111\u1ebfn v\u1ecb tr\u00ed cao nh\u1ea5t th\u00ec kho\u1ea3ng c\u00e1ch gi\u1eefa hai v\u1eadt b\u1eb1ng bao nhi\u00eau? Bi\u1ebft r\u1eb1ng \u0111\u1ed9 cao \u0111\u1ee7 l\u1edbn.<\/p>\n A.<\/strong>20 cm. B.<\/strong>50 cm. C.<\/strong>80 cm. D.<\/strong>70 cm.<\/p>\n C\u00e2u 21:<\/strong>S\u00f3ng truy\u1ec1n v\u1edbi t\u1ed1c \u0111\u1ed9 kh\u00f4ng \u0111\u1ed5i 10m\/s t\u1eeb \u0111i\u1ec3m M \u0111\u1ebfn O tr\u00ean c\u00f9ng ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng v\u1edbi MO = 50cm, coi bi\u00ean \u0111\u1ed9 s\u00f3ng kh\u00f4ng \u0111\u1ed5i. Bi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh s\u00f3ng t\u1ea1i O l\u00e0 u<\/p>\n = 5cos(10\u03c0t) cm. Ph\u01b0\u01a1ng tr\u00ecnh s\u00f3ng t\u1ea1i M l\u00e0:<\/p>\n A.<\/strong>u = 5cos(10\u03c0t – \u03c0\/4) cm B.<\/strong>u = 5cos(10\u03c0t – \u03c0\/2) cm<\/p>\n C.<\/strong>u = 5cos(10\u03c0t + \u03c0\/6) cm D.<\/strong>u = 5cos(10\u03c0t + \u03c0\/2) cm<\/p>\n C\u00e2u 22:<\/strong>B\u01b0\u1edbc s\u00f3ng l\u00e0 kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111i\u1ec3m<\/p>\n A.<\/strong>g\u1ea7n nhau nh\u1ea5t m\u00e0 dao \u0111\u1ed9ng t\u1ea1i hai \u0111i\u1ec3m \u0111\u00f3 c\u00f9ng pha.<\/p>\n B.<\/strong>g\u1ea7n nhau nh\u1ea5t tr\u00ean c\u00f9ng m\u1ed9t ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng m\u00e0 dao \u0111\u1ed9ng t\u1ea1i hai \u0111i\u1ec3m \u0111\u00f3 c\u00f9ng pha.<\/p>\n C.<\/strong>tr\u00ean c\u00f9ng m\u1ed9t ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng m\u00e0 dao \u0111\u1ed9ng t\u1ea1i hai \u0111i\u1ec3m \u0111\u00f3 ng\u01b0\u1ee3c pha.<\/p>\n D.<\/strong>tr\u00ean c\u00f9ng m\u1ed9t ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng m\u00e0 dao \u0111\u1ed9ng t\u1ea1i hai \u0111i\u1ec3m \u0111\u00f3 c\u00f9ng pha.<\/p>\n C\u00e2u 23:<\/strong>M\u1ed9t s\u1ee3i d\u00e2y \u0111\u00e0n h\u1ed3i AB d\u00e0i 2m \u0111\u01b0\u1ee3c c\u0103ng ngang, \u0111\u1ea7u B c\u1ed1 \u0111\u1ecbnh, \u0111\u1ea7u A g\u1eafn v\u00e0o 1 c\u1ea7n rung dao \u0111\u1ed9ng v\u1edbi t\u1ea7n s\u1ed1 50Hz. T\u1ed1c \u0111\u1ed9 truy\u1ec1n s\u00f3ng tr\u00ean d\u00e2y l\u00e0 50m\/s. Tr\u00ean d\u00e2y h\u00ecnh th\u00e0nh s\u00f3ng d\u1eebng v\u1edbi<\/p>\n A.<\/strong>3 b\u1ee5ng, 4 n\u00fat. B.<\/strong>2 b\u1ee5ng, 3 n\u00fat. C.<\/strong>4 b\u1ee5ng, 5 n\u00fat. D.<\/strong>1 b\u1ee5ng, 2 n\u00fat.<\/p>\n C\u00e2u 24:<\/strong>Con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a v\u1edbi t\u1ea7n s\u1ed1 f. \u0110\u1ed9ng n\u0103ng v\u00e0 th\u1ebf n\u0103ng c\u1ee7a con l\u1eafc bi\u1ebfn thi\u00ean tu\u1ea7n ho\u00e0n v\u1edbi t\u1ea7n s\u1ed1 l\u00e0<\/p>\n A.<\/strong>2f. B.<\/strong>f. C.<\/strong>4f. D.<\/strong>f\/2.<\/p>\n C\u00e2u 25:<\/strong>Hai \u00e2m c\u00f9ng \u0111\u1ed9 cao l\u00e0 hai \u00e2m c\u00f3 c\u00f9ng<\/p>\n A.<\/strong>bi\u00ean \u0111\u1ed9. B.<\/strong>t\u1ea7n s\u1ed1. C.<\/strong>c\u01b0\u1eddng \u0111\u1ed9 \u00e2m. D.<\/strong>m\u1ee9c c\u01b0\u1eddng \u0111\u1ed9 \u00e2m.<\/p>\n C\u00e2u 26:<\/strong>Nh\u1eadn \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0sai<\/u><\/strong>khi n\u00f3i v\u1ec1 dao \u0111\u1ed9ng c\u01a1 h\u1ecdc t\u1eaft d\u1ea7n?<\/p>\n A.<\/strong>Dao \u0111\u1ed9ng t\u1eaft d\u1ea7n l\u00e0 dao \u0111\u1ed9ng c\u00f3 bi\u00ean \u0111\u1ed9 gi\u1ea3m d\u1ea7n theo th\u1eddi gian.<\/p>\n B.<\/strong>Trong dao \u0111\u1ed9ng t\u1eaft d\u1ea7n, c\u01a1 n\u0103ng gi\u1ea3m d\u1ea7n theo th\u1eddi gian.<\/p>\n C.<\/strong>L\u1ef1c ma s\u00e1t c\u00e0ng l\u1edbn th\u00ec dao \u0111\u1ed9ng t\u1eaft c\u00e0ng nhanh.<\/p>\n D.<\/strong>Dao<\/u>\u0111\u1ed9ng t\u1eaft d\u1ea7n c\u00f3 \u0111\u1ed9ng n\u0103ng gi\u1ea3m d\u1ea7n c\u00f2n th\u1ebf n\u0103ng bi\u1ebfn thi\u00ean \u0111i\u1ec1u h\u00f2a<\/p>\n C\u00e2u 27:<\/strong>M\u1ed9t con l\u1eafc \u0111\u01a1n d\u00e0i l = 25cm, h\u00f2n bi c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng m = 10g v\u00e0 mang \u0111i\u1ec7n t\u00edch q = 10<\/p>\n C. Treo con l\u1eafc v\u00e0o gi\u1eefa hai b\u1ea3n kim lo\u1ea1i th\u1eb3ng \u0111\u1ee9ng, song song c\u00e1ch nhau d = 22cm. \u0110\u1eb7t v\u00e0o hai b\u1ea3n hi\u1ec7u \u0111i\u1ec7n th\u1ebf m\u1ed9t chi\u1ec1u U = 88V, l\u1ea5y g = 10 m\/s<\/p>\n . Chu k\u00ec dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a v\u1edbi bi\u00ean \u0111\u1ed9 nh\u1ecf c\u1ee7a con l\u1eafc \u0111\u01a1n l\u00e0:<\/p>\n A.<\/strong>0,659s B.<\/strong>0,983 s C.<\/strong>0,897s D.<\/strong>0,957 s<\/p>\n C\u00e2u 28:<\/strong>Hai ngu\u1ed3n s\u00f3ng c\u01a1 k\u1ebft h\u1ee3p S<\/p>\n , S<\/p>\n dao \u0111\u1ed9ng c\u00f9ng ph\u01b0\u01a1ng, c\u00f9ng bi\u00ean \u0111\u1ed9 v\u00e0 ng\u01b0\u1ee3c pha c\u00e1ch nhau 60 cm c\u00f3 t\u1ea7n s\u1ed1 5 Hz, t\u1ed1c \u0111\u1ed9 truy\u1ec1n s\u00f3ng l\u00e0 40 cm\/s. S\u1ed1 c\u1ef1c ti\u1ec3u giao thoa tr\u00ean \u0111o\u1ea1n S<\/p>\n S<\/p>\n l\u00e0:<\/p>\n A.<\/strong>15. B.<\/strong>16. C.<\/strong>14. D.<\/strong>13.<\/p>\n C\u00e2u 29:<\/strong>Con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0, khi t\u0103ng kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a v\u1eadt l\u00ean 4 l\u1ea7n th\u00ec chu k\u00ec dao \u0111\u1ed9ng c\u1ee7a v\u1eadt<\/p>\n A.<\/strong>t\u0103ng l\u00ean 2 l\u1ea7n. B.<\/strong>gi\u1ea3m \u0111i 2 l\u1ea7n. C.<\/strong>gi\u1ea3m \u0111i 4 l\u1ea7n. D.<\/strong>t\u0103ng l\u00ean 4 l\u1ea7n.<\/p>\n C\u00e2u 30:<\/strong>Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng c\u1ee7a m\u1ed9t v\u1eadt c\u00f3 d\u1ea1ng x = Asin(t +\/4). Ch\u1ecdn k\u1ebft lu\u1eadn\u0111\u00fang<\/em>.<\/p>\n A.<\/strong>V\u1eadt dao \u0111\u1ed9ng v\u1edbi bi\u00ean \u0111\u1ed9 A. B.<\/strong>V\u1eadt dao \u0111\u1ed9ng v\u1edbi pha ban \u0111\u1ea7u\/4.<\/p>\n C.<\/strong>V\u1eadt dao \u0111\u1ed9ng v\u1edbi bi\u00ean \u0111\u1ed9 2A. D.<\/strong>V\u1eadt dao \u0111\u1ed9ng v\u1edbi bi\u00ean \u0111\u1ed9 A\/2.<\/p>\n C\u00e2u 31:<\/strong>Ngu\u1ed3n \u00e2m S ph\u00e1t ra \u00e2m c\u00f3 c\u00f4ng su\u1ea5t kh\u00f4ng \u0111\u1ed5i truy\u1ec1n \u0111\u1eb3ng h\u01b0\u1edbng v\u1ec1 m\u1ecdi ph\u01b0\u01a1ng. T\u1ea1i M c\u00e1ch S \u0111o\u1ea1n 2m, m\u1ee9c c\u01b0\u1eddng \u0111\u1ed9 \u00e2m l\u00e0 50 dB. Gi\u1ea3 s\u1eed m\u00f4i tr\u01b0\u1eddng kh\u00f4ng h\u1ea5p th\u1ee5 \u00e2m. M\u1ee9c c\u01b0\u1eddng \u0111\u1ed9 \u00e2m t\u1ea1i N c\u00e1ch S \u0111o\u1ea1n 8m l\u00e0:<\/p>\n A.<\/strong>38 dB. B.<\/strong>48 dB. C.<\/strong>42 dB. D.<\/strong>45 dB.<\/p>\n T\u1ea5t c\u1ea3 n\u1ed9i dung b\u00e0i vi\u1ebft. C\u00e1c em h\u00e3y xem th\u00eam v\u00e0 t\u1ea3i file chi ti\u1ebft t\u1ea1i \u0111\u00e2y:<\/strong>Download<\/p>\n","protected":false},"excerpt":{"rendered":" Tr\u01b0\u1eddng THPT Ho\u00e0ng L\u1ec7 Kha, Thanh H\u00f3a t\u1ed5 ch\u1ee9c thi th\u1eed THPT Qu\u1ed1c gia n\u0103m 2016 m\u00f4n L\u00fd m\u00e3 \u0111\u1ec1 132, c\u00f3 \u0111\u00e1p \u00e1n, xem chi ti\u1ebft \u0111\u1ec1 thi v\u00e0 \u0111\u00e1p \u00e1n d\u01b0\u1edbi \u0111\u00e2y: \u0110\u1ec1 thi th\u1eed THPTQG m\u00f4n L\u00fd THPT Ho\u00e0ng L\u1ec7 Kha 2016 C\u00e2u 14:S\u00f3ng \u0111i\u1ec7n t\u1eeb A.l\u00e0 s\u00f3ng d\u1ecdc ho\u1eb7c s\u00f3ng ngang. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[25],"tags":[],"yoast_head":"\n<\/strong><\/p>\n<\/p>\n\u0110\u00e1p \u00e1n \u0111\u1ec1 thi th\u1eed THPTQG m\u00f4n L\u00fd THPT Ho\u00e0ng L\u1ec7 Kha 2016<\/strong><\/h3>\n
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\n 1<\/td>\n C<\/td>\n 16<\/td>\n A<\/td>\n 31<\/td>\n A<\/td>\n 46<\/td>\n B<\/td>\n<\/tr>\n \n 2<\/td>\n D<\/td>\n 17<\/td>\n D<\/td>\n 32<\/td>\n B<\/td>\n 47<\/td>\n B<\/td>\n<\/tr>\n \n 3<\/td>\n A<\/td>\n 18<\/td>\n C<\/td>\n 33<\/td>\n C<\/td>\n 48<\/td>\n D<\/td>\n<\/tr>\n \n 4<\/td>\n C<\/td>\n 19<\/td>\n D<\/td>\n 34<\/td>\n C<\/td>\n 49<\/td>\n D<\/td>\n<\/tr>\n \n 5<\/td>\n B<\/td>\n 20<\/td>\n C<\/td>\n 35<\/td>\n B<\/td>\n 50<\/td>\n C<\/td>\n<\/tr>\n \n 6<\/td>\n B<\/td>\n 21<\/td>\n D<\/td>\n 36<\/td>\n D<\/td>\n<\/tr>\n \n 7<\/td>\n B<\/td>\n 22<\/td>\n B<\/td>\n 37<\/td>\n D<\/td>\n<\/tr>\n \n 8<\/td>\n C<\/td>\n 23<\/td>\n C<\/td>\n 38<\/td>\n C<\/td>\n<\/tr>\n \n 9<\/td>\n B<\/td>\n 24<\/td>\n A<\/td>\n 39<\/td>\n A<\/td>\n<\/tr>\n \n 10<\/td>\n B<\/td>\n 25<\/td>\n B<\/td>\n 40<\/td>\n B<\/td>\n<\/tr>\n \n 11<\/td>\n D<\/td>\n 26<\/td>\n D<\/td>\n 41<\/td>\n A<\/td>\n<\/tr>\n \n 12<\/td>\n D<\/td>\n 27<\/td>\n D<\/td>\n 42<\/td>\n D<\/td>\n<\/tr>\n \n 13<\/td>\n A<\/td>\n 28<\/td>\n A<\/td>\n 43<\/td>\n C<\/td>\n<\/tr>\n \n 14<\/td>\n B<\/td>\n 29<\/td>\n A<\/td>\n 44<\/td>\n C<\/td>\n<\/tr>\n \n 15<\/td>\n A<\/td>\n 30<\/td>\n A<\/td>\n 45<\/td>\n A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
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