Ch\u01b0\u01a1ng 3 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp ch\u01b0\u01a1ng 3 (Ph\u1ea7n 3)<\/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\/chuong-3-bai-tap-trac-nghiem-giai-tich-12-on-tap-chuong-3-phan-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Ch\u01b0\u01a1ng 3 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp ch\u01b0\u01a1ng 3 (Ph\u1ea7n 3)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 13:\u00a0Di\u1ec7n t\u00edch h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi \u0111\u01b0\u1eddng cong: y = x2\u00a0+ 1 , ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u01b0\u1eddng cong n\u00e0y t\u1ea1i M(2;5) v\u00e0 tr\u1ee5c Oy l\u00e0: A. 0 \u00a0\u00a0\u00a0 B. -8\/3 \u00a0\u00a0\u00a0 C. 8\/3 \u00a0\u00a0\u00a0 D. 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{"id":26888,"date":"2018-04-16T11:56:40","date_gmt":"2018-04-16T04:56:40","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=26888"},"modified":"2018-04-16T11:56:40","modified_gmt":"2018-04-16T04:56:40","slug":"chuong-3-bai-tap-trac-nghiem-giai-tich-12-on-tap-chuong-3-phan-3","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/chuong-3-bai-tap-trac-nghiem-giai-tich-12-on-tap-chuong-3-phan-3\/","title":{"rendered":"Ch\u01b0\u01a1ng 3 – B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp ch\u01b0\u01a1ng 3 (Ph\u1ea7n 3)"},"content":{"rendered":"

C\u00e2u 13:<\/b>\u00a0Di\u1ec7n t\u00edch h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi \u0111\u01b0\u1eddng cong: y = x^{2<\/sup>\u00a0+ 1 , ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u01b0\u1eddng cong n\u00e0y t\u1ea1i M(2;5) v\u00e0 tr\u1ee5c Oy l\u00e0:<\/p>\n}

A. 0 \u00a0\u00a0\u00a0 B. -8\/3 \u00a0\u00a0\u00a0 C. 8\/3 \u00a0\u00a0\u00a0 D. K\u1ebft qu\u1ea3 kh\u00e1c .<\/p>\n

C\u00e2u 14:<\/b>\u00a0Th\u1ec3 t\u00edch v\u1eadt th\u1ec3 tr\u00f2n xoay sinh ra b\u1edfi ph\u00e9p quay quanh tr\u1ee5c Ox c\u1ee7a h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi tr\u1ee5c Ox v\u00e0 y = \u221axsinx v\u1edbi (0 \u2264 x \u2264 \u03c0) l\u00e0:<\/p>\n

$\"\"$ <\/p>\n

C\u00e2u 15:<\/b>\u00a0T\u00ednh th\u1ec3 t\u00edch v\u1eadt th\u1ec3 tr\u00f2n xoay quanh tr\u1ee5c Ox sinh b\u1edfi h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi c\u00e1c \u0111\u01b0\u1eddng<\/p>\n

$\"B\u00e0i$ $\"\"$ <\/p>\n

C\u00e2u 16:<\/b>\u00a0T\u00ednh th\u1ec3 t\u00edch v\u1eadt th\u1ec3 tr\u00f2n xoay quanh tr\u1ee5c Oy sinh b\u1edfi h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi c\u00e1c \u0111\u01b0\u1eddng y = 2, y = 4 , y = x^{2<\/sup>\/2 .<\/p>\n}

A. 12\u03c0 \u00a0\u00a0\u00a0 B. -12\u03c0\u00a0\u00a0\u00a0 C. 16\u03c0\u00a0\u00a0\u00a0 D. -16\u03c0<\/p>\n

C\u00e2u 17:<\/b>\u00a0Th\u1ec3 t\u00edch v\u1eadt th\u1ec3 tr\u00f2n xoay khi quay h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi c\u00e1c \u0111\u01b0\u1eddng y = tanx, y = 0, x = 0, x = \u03c0\/3 quanh Ox l\u00e0:<\/p>\n

$\"\"$ <\/p>\n

C\u00e2u 18:<\/b>\u00a0Di\u1ec7n t\u00edch h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi ay = x_{2<\/sub>\u00a0v\u00e0 ax = y_{2<\/sub>\u00a0l\u00e0:<\/p>\n}}

A. -a^{3<\/sup>\/3\u00a0\u00a0\u00a0 B. a^{3<\/sup>\/3 \u00a0\u00a0\u00a0 C. a^{2<\/sup>\u00a0\u00a0\u00a0 D. -a^{2<\/sup><\/p>\n}}}}

C\u00e2u 19:<\/b>\u00a0M\u1ed9t v\u1eadt chuy\u1ec3n \u0111\u1ed9ng v\u1edbi v\u1eadn t\u1ed1c<\/p>\n

$\"\"$ <\/p>\n

Qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt \u0111i \u0111\u01b0\u1ee3c sau 4s x\u1ea5p x\u1ec9 b\u1eb1ng:<\/p>\n

A. 11m \u00a0\u00a0\u00a0 B. 12m \u00a0\u00a0\u00a0C. 13m \u00a0\u00a0\u00a0 D. 14m.<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
13-C<\/td>\n 14-B<\/td>\n 15-D<\/td>\n 16-A<\/td>\n 17-D<\/td>\n 18-B<\/td>\n 19-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
C\u00e2u 13:<\/b><\/p>\n
Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn v\u1edbi y = x^{2<\/sup>\u00a0+ 1 t\u1ea1i M(2;5) l\u00e0: y = 4(x – 2) + 5 = 4x – 3.<\/p>\n}
Ta c\u00f3 x^{2<\/sup>\u00a0+ 1 = 4x – 3 => x = 2 khi \u0111\u00f3<\/p>\n}
$\"\"$ <\/p>\n
C\u00e2u 14:<\/b><\/p>\n
$\"\"$ <\/p>\n
C\u00e2u 15:<\/b><\/p>\n
$\"\"$ <\/p>\n
C\u00e2u 16:<\/b><\/p>\n
$\"\"$ <\/p>\n
C\u00e2u 17:<\/b><\/p>\n
$\"\"$ <\/p>\n
C\u00e2u 18:<\/b><\/p>\n
$\"\"$ <\/p>\n
C\u00e2u 19:<\/b><\/p>\n
Qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt di chuy\u1ec3n sau th\u1eddi gian 4 gi\u00e2y b\u1eb1ng :<\/p>\n
$\"\"$ <\/p>\n","protected":false},"excerpt":{"rendered":"
C\u00e2u 13:\u00a0Di\u1ec7n t\u00edch h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi \u0111\u01b0\u1eddng cong: y = x2\u00a0+ 1 , ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u01b0\u1eddng cong n\u00e0y t\u1ea1i M(2;5) v\u00e0 tr\u1ee5c Oy l\u00e0: A. 0 \u00a0\u00a0\u00a0 B. -8\/3 \u00a0\u00a0\u00a0 C. 8\/3 \u00a0\u00a0\u00a0 D. K\u1ebft qu\u1ea3 kh\u00e1c . C\u00e2u 14:\u00a0Th\u1ec3 t\u00edch v\u1eadt th\u1ec3 tr\u00f2n xoay sinh ra b\u1edfi ph\u00e9p quay quanh tr\u1ee5c […]<\/p>\n","protected":false},"author":3,"featured_media":26889,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1447,1446],"yoast_head":"\nCh\u01b0\u01a1ng 3 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp ch\u01b0\u01a1ng 3 (Ph\u1ea7n 3)<\/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\/chuong-3-bai-tap-trac-nghiem-giai-tich-12-on-tap-chuong-3-phan-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Ch\u01b0\u01a1ng 3 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp ch\u01b0\u01a1ng 3 (Ph\u1ea7n 3)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 13:\u00a0Di\u1ec7n t\u00edch h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi \u0111\u01b0\u1eddng cong: y = x2\u00a0+ 1 , ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u01b0\u1eddng cong n\u00e0y t\u1ea1i M(2;5) v\u00e0 tr\u1ee5c Oy l\u00e0: A. 0 \u00a0\u00a0\u00a0 B. -8\/3 \u00a0\u00a0\u00a0 C. 8\/3 \u00a0\u00a0\u00a0 D. 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