\n| 1-C<\/td>\n | 2-C<\/td>\n | 3-D<\/td>\n | 4-C<\/td>\n | 5-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n C\u00e2u 1:<\/b><\/p>\n Ta c\u00f3: a = 2 > 0; y’ = 6x2<\/sup>\u00a0– 6(m + 1)x + 6(m + 1)2<\/sup>\u00a0= 6[x2<\/sup>\u00a0– (m + 1)x + (m + 1)2<\/sup>]<\/p>\n\u0394 = -3(m + 1)2<\/sup>\u00a0\u2264 0 \u2200x \u2208 R => y’ = 0 v\u00f4 nghi\u1ec7m ho\u1eb7c nghi\u1ec7m k\u00e9p<\/p>\nC\u00e2u 2:<\/b><\/p>\n Ta c\u00f3 a = 1 > 0; => y\u2019= 0 c\u00f3 m\u1ed9t nghi\u1ec7m x = 0.<\/p>\n C\u00e2u 4:<\/b><\/p>\n \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng y= 3, y=-4 t\u1ea1i m\u1ed9t \u0111i\u1ec3m ; c\u1eaft tr\u1ee5c ho\u00e0nh v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y = 5\/3 t\u1ea1i 3 \u0111i\u1ec3m.<\/p>\n \n <\/div>\n<\/div>\n C\u00e2u 5:<\/b><\/p>\n y’ = 3x2<\/sup>\u00a0. \u0110\u01b0\u1eddng th\u1eb3ng y = 3x + m l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng cong y = x3<\/sup>\u00a0+ 2<\/p>\n![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/4-128.png\") <\/p>\n
Ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng cong t\u1ea1i A(1;3) l\u00e0 y=3x.<\/p>\n Ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng cong t\u1ea1i B(-1;1) l\u00e0 y=3x+4.<\/p>\n Do \u0111\u00f3 m \u2208 {0; 4}<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Cho h\u00e0m s\u1ed1 2×3\u00a0– 3(m+1)x2\u00a0+ 6(m + 1)2x + 1. H\u00ecnh n\u00e0o d\u01b0\u1edbi \u0111\u00e2y m\u00f4 t\u1ea3 ch\u00ednh x\u00e1c nh\u1ea5t \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 tr\u00ean? C\u00e2u 2:\u00a0Cho h\u00e0m s\u1ed1 y = x4\u00a0+ (m2\u00a0+ 1)x2\u00a0+ 1. H\u00ecnh n\u00e0o d\u01b0\u1edbi \u0111\u00e2y m\u00f4 t\u1ea3 ch\u00ednh x\u00e1c nh\u1ea5t \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 tr\u00ean? C\u00e2u 3:\u00a0Cho \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 […]<\/p>\n","protected":false},"author":3,"featured_media":27888,"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":"\n Ch\u01b0\u01a1ng 1 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (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-1-bai-tap-trac-nghiem-giai-tich-12-khao-sat-su-bien-thien-va-ve-do-thi-ham-so-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 1 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (Ph\u1ea7n 3)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 1:\u00a0Cho h\u00e0m s\u1ed1 2x3\u00a0– 3(m+1)x2\u00a0+ 6(m + 1)2x + 1. 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