C\u00e2u 8:<\/b>\u00a0S\u00f4\u0301 \u0111i\u00ea\u0309m c\u01b0\u0323c ti\u00ea\u0309u cu\u0309a ha\u0300m s\u00f4\u0301 y = x4<\/sup>\u00a0+ x2<\/sup>\u00a0+ 1 la\u0300<\/p>\n A. 0 \u00a0\u00a0\u00a0B. 1 \u00a0\u00a0\u00a0 C. 2\u00a0\u00a0\u00a0D. 3.<\/p>\n C\u00e2u 9:<\/b>\u00a0Cho ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 1. Ti\u0301ch ca\u0301c gia\u0301 tri\u0323 c\u01b0\u0323c \u0111a\u0323i cu\u0309a ha\u0300m s\u00f4\u0301 la\u0300<\/p>\n A. 0 \u00a0\u00a0\u00a0B. \u20133 \u00a0\u00a0\u00a0C. 3\u00a0\u00a0\u00a0D. \u20136<\/p>\n C\u00e2u 10:<\/b>\u00a0\u0110\u01b0\u01a1\u0300ng th\u0103\u0309ng \u0111i qua hai \u0111i\u00ea\u0309m c\u01b0\u0323c tri\u0323 cu\u0309a ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 2 co\u0301 ph\u01b0\u01a1ng tri\u0300nh la\u0300<\/p>\n A. y = -x + 2 \u00a0\u00a0\u00a0B. y = x + 2\u00a0\u00a0\u00a0C. y = 2x + 2\u00a0\u00a0\u00a0D. y = -2x + 2<\/p>\n C\u00e2u 11:<\/b>\u00a0Ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 6x2<\/sup>\u00a0co\u0301 gia\u0301 tri\u0323 nho\u0309 nh\u00e2\u0301t va\u0300 gia\u0301 tri\u0323 l\u01a1\u0301n nh\u00e2\u0301t tr\u00ean \u0111oa\u0323n [-1; 5] t\u01b0\u01a1ng \u01b0\u0301ng la\u0300<\/p>\n A. \u201325 va\u0300 \u20137 \u00a0\u00a0\u00a0B. \u20137 va\u0300 0 \u00a0\u00a0\u00a0C. \u201332 va\u0300 0 \u00a0\u00a0\u00a0D. \u201332 va\u0300 \u20137.<\/p>\n C\u00e2u 12:<\/b>\u00a0Ti\u00ea\u0301p tuy\u00ea\u0301n ta\u0323i \u0111i\u00ea\u0309m A(0; 2) cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x + 2 co\u0301 ph\u01b0\u01a1ng tri\u0300nh la\u0300<\/p>\n A. y = -3x + 2\u00a0\u00a0\u00a0B. y = 3x + 2\u00a0\u00a0\u00a0C. y = 2x + 2\u00a0\u00a0\u00a0D. y = x + 2<\/p>\n C\u00e2u 13:<\/b>\u00a0Cho ha\u0300m s\u00f4\u0301 y = 2x4<\/sup>\u00a0– 5x2<\/sup>\u00a0– 7. S\u00f4\u0301 ti\u00ea\u0301p tuy\u00ea\u0301n \u0111i qua \u0111i\u00ea\u0309m M(0; -7) cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 la\u0300<\/p>\n A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0 C. 3 \u00a0\u00a0\u00a0D. 4<\/p>\n C\u00e2u 14:<\/b>\u00a0S\u00f4\u0301 giao \u0111i\u00ea\u0309m cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 2x + 1 v\u01a1\u0301i tru\u0323c hoa\u0300nh la\u0300<\/p>\n A. 0\u00a0\u00a0\u00a0B. 1\u00a0\u00a0\u00a0C. 2\u00a0\u00a0\u00a0D. 4<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 8:<\/b><\/p>\n Ta c\u00f3: y’ = 4x3<\/sup>\u00a0+ 2x = 2x(2x2<\/sup>\u00a0+ 1) . Do \u0111\u00f3: y’ = 0 <=> x = 0 L\u1eadp b\u1ea3ng bi\u1ebfn thi\u00ean ta suy ra h\u00e0m s\u1ed1 \u0111\u00e3 cho ch\u1ec9 c\u00f3 m\u1ed9t c\u1ef1c ti\u1ec3u x = 0<\/p>\n C\u00e2u 9:<\/b><\/p>\n Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 6x; y’ = 0 <=> x = 0, x = 2. C\u00e1c gi\u00e1 tr\u1ecb c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 l\u00e0:<\/p>\n y1<\/sub>\u00a0= y(0) = 1, y2<\/sub>\u00a0= y(2) = -3<\/p>\n V\u1eady t\u00edch c\u00e1c gi\u00e1 tr\u1ecb c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 y1<\/sub>y2<\/sub>\u00a0= -3<\/p>\n C\u00e2u 10:<\/b><\/p>\n Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 6x, y’ = 0 <=> x = 0, x = 2. C\u00e1c \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0 A(0; 2), B(2; -2)<\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng AB l\u00e0<\/p>\n <\/p>\n C\u00e2u 11:<\/b><\/p>\n Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 12x, y’ = 0 <=> x = 0, x = 4 So s\u00e1nh c\u00e1c gi\u00e1 tr\u1ecb:<\/p>\n y(-1) = -7, y(0) = 0, y(4) = -32, y(5) = -25<\/p>\n Ta c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho tr\u00ean \u0111o\u1ea1n [-1; 5] l\u00e0 0 v\u00e0 -32.<\/p>\n C\u00e2u 12:<\/b><\/p>\n Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 3. H\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a ti\u1ebfp tuy\u1ebfn l\u00e0: k = y'(0) = -3<\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn l\u00e0 y – 2 = -3(x – 0) <=> y = -3x + 2<\/p>\n C\u00e2u 13:<\/b><\/p>\n X\u00e9t \u0111i\u1ec3m A(x0<\/sub>, y0<\/sub>) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn d t\u1ea1i A c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 l\u00e0<\/p>\n <\/p>\n Ti\u1ebfp tuy\u1ebfn \u0111i qua M(0;-7) khi v\u00e0 ch\u1ec9 khi<\/p>\n <\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean (\u1ea9n x0<\/sub>) c\u00f3 ba nghi\u1ec7m n\u00ean c\u00f3 ba ti\u1ebfp tuy\u1ebfn th\u1ecfa m\u00e3n y\u00eau c\u1ea7u b\u00e0i to\u00e1n.<\/p>\n C\u00e2u 14:<\/b><\/p>\n Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 6x + 2,<\/p>\n <\/p>\n Gi\u00e1 tr\u1ecb c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 l\u00e0:<\/p>\n <\/p>\n L\u1eadp b\u1ea3ng bi\u1ebfn thi\u00ean ta suy ra \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u1eaft tr\u1ee5c ho\u00e0nh t\u1ea1i 1 \u0111i\u1ec3m.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 8:\u00a0S\u00f4\u0301 \u0111i\u00ea\u0309m c\u01b0\u0323c ti\u00ea\u0309u cu\u0309a ha\u0300m s\u00f4\u0301 y = x4\u00a0+ x2\u00a0+ 1 la\u0300 A. 0 \u00a0\u00a0\u00a0B. 1 \u00a0\u00a0\u00a0 C. 2\u00a0\u00a0\u00a0D. 3. C\u00e2u 9:\u00a0Cho ha\u0300m s\u00f4\u0301 y = x3\u00a0– 3×2\u00a0+ 1. Ti\u0301ch ca\u0301c gia\u0301 tri\u0323 c\u01b0\u0323c \u0111a\u0323i cu\u0309a ha\u0300m s\u00f4\u0301 la\u0300 A. 0 \u00a0\u00a0\u00a0B. \u20133 \u00a0\u00a0\u00a0C. 3\u00a0\u00a0\u00a0D. \u20136 C\u00e2u 10:\u00a0\u0110\u01b0\u01a1\u0300ng th\u0103\u0309ng \u0111i qua hai \u0111i\u00ea\u0309m […]<\/p>\n","protected":false},"author":3,"featured_media":26595,"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\n\n
\n 8-B<\/td>\n 9-B<\/td>\n 10-D<\/td>\n 11-C<\/td>\n 12-A<\/td>\n 13-C<\/td>\n 14-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n