C\u00e1c em tham kh\u1ea3o \u0110\u1ec1 thi gi\u1eefa h\u1ecdc k\u00ec 1 m\u00f4n To\u00e1n l\u1edbp 11 n\u0103m 2015 THPT Nguy\u1ec5n Trung Tr\u1ef1c tham kh\u1ea3o chi ti\u1ebft d\u01b0\u1edbi \u0111\u00e2y:<\/p>\n
C\u00e2u 1 : (5 \u0111i\u1ec3m)<\/strong>\u00a0\u00a0\u00a0 Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau:<\/p>\n <\/p>\n C\u00e2u 2 : (1 \u0111i\u1ec3m)<\/strong> T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1:<\/p>\n y = 4 \u2013 3sin x<\/p>\n C\u00e2u 3 : (1 \u0111i\u1ec3m)<\/strong><\/p>\n Trong m\u1eb7t ph\u1eb3ng Oxy cho \u0111i\u1ec3m A(-1; 2), v\u00e9c t\u01a1 v = (3;5) v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) c\u00f3 \u00a0\u00a0\u00a0ph\u01b0\u01a1ng tr\u00ecnh: \u00a03x \u2013 5y +3 = 0.<\/p>\n a)\u00a0\u00a0\u00a0\u00a0 T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m A\u2019 l\u00e0 \u1ea3nh c\u1ee7a \u0111i\u1ec3m A qua ph\u00e9p t\u1ecbnh ti\u1ebfn theo vecto v .<\/p>\n b) T\u00ecm ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng (d\u2019) l\u00e0 \u1ea3nh c\u1ee7a (d) qua ph\u00e9p t\u1ecbnh ti\u1ebfn theo vecto v<\/p>\n C\u00e2u 4 : (3 \u0111i\u1ec3m)<\/strong><\/p>\n \u00a0\u00a0\u00a0\u00a0 <\/strong>Cho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 \u0111\u00e1y ABCD l\u00e0 h\u00ecnh thang (AD \/\/ BC). G\u1ecdi M, N, K l\u1ea7n l\u01b0\u1ee3t l\u00e0 c\u00e1c \u0111i\u1ec3m tr\u00ean c\u1ea1nh SA, SB, SD sao cho\u00a0<\/p>\n a)\u00a0\u00a0\u00a0\u00a0 X\u00e1c \u0111\u1ecbnh giao tuy\u1ebfn c\u1ee7a (SAB) v\u00e0 (SCD).<\/p>\n b)\u00a0\u00a0\u00a0\u00a0 X\u00e1c \u0111\u1ecbnh giao \u0111i\u1ec3m I c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng NK v\u00e0 m\u1eb7t ph\u1eb3ng (SAC).<\/p>\n c)\u00a0\u00a0\u00a0\u00a0 G\u1ecdi E = MN \u2229 AB; \u00a0F = MK \u2229 AD; \u00a0Q = MI \u2229 AC.<\/p>\n Ch\u1ee9ng minh 3 \u0111i\u1ec3m E, F, Q \u00a0th\u1eb3ng h\u00e0ng.<\/p>\n —— H\u1ebft —–\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong><\/p>\n <\/p>\n Th\u00ed sinh kh\u00f4ng \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng t\u00e0i li\u1ec7u. C\u00e1n b\u1ed9 coi thi kh\u00f4ng gi\u1ea3i th\u00edch g\u00ec th\u00eam.<\/strong><\/p>\n \u0110\u00e1p \u00e1n<\/strong><\/span><\/p>\n <\/strong><\/span><\/p>\n <\/strong><\/span><\/p>\n <\/strong><\/span><\/p>\n <\/strong><\/span><\/p>\n <\/strong><\/span><\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e1c em tham kh\u1ea3o \u0110\u1ec1 thi gi\u1eefa h\u1ecdc k\u00ec 1 m\u00f4n To\u00e1n l\u1edbp 11 n\u0103m 2015 THPT Nguy\u1ec5n Trung Tr\u1ef1c tham kh\u1ea3o chi ti\u1ebft d\u01b0\u1edbi \u0111\u00e2y: C\u00e2u 1 : (5 \u0111i\u1ec3m)\u00a0\u00a0\u00a0 Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau: C\u00e2u 2 : (1 \u0111i\u1ec3m) T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1: y […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[171],"tags":[],"yoast_head":"\n
\n<\/strong><\/span><\/p>\n
\n<\/strong><\/span><\/p>\n