C\u00e2u 7:<\/b>\u00a0\u0110\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n
<\/p>\n
A.y = 1 \u00a0\u00a0\u00a0 B. y = 0 \u00a0\u00a0\u00a0 C. y = -1 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n
C\u00e2u 8:<\/b>\u00a0\u0110\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n
<\/p>\n
A. x = 0 \u00a0\u00a0\u00a0 B. x = 2, x = -2 \u00a0\u00a0\u00a0 C. x – 2 = 0 \u00a0\u00a0\u00a0 D. x + 2 = 0<\/p>\n
C\u00e2u 9:<\/b>\u00a0Cho h\u00e0m s\u1ed1<\/p>\n
<\/p>\n
H\u1ecfi giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 tr\u00ean lu\u00f4n n\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng c\u1ed1 \u0111\u1ecbnh c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh n\u00e0o trong c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau?<\/p>\n
A. y = x\u00a0\u00a0\u00a0 B. x2<\/sup>\u00a0+ y2<\/sup>\u00a0= 1 \u00a0\u00a0\u00a0 C. y = x2<\/sup>\u00a0\u00a0\u00a0\u00a0 D. y = x3<\/sup><\/p>\n C\u00e2u 10:<\/b>\u00a0\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– mx2<\/sup>\u00a0+ 2 c\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn?<\/p>\n A. 1 \u00a0\u00a0\u00a0 B. 2 \u00a0\u00a0\u00a0C. 0 \u00a0\u00a0\u00a0 D. 3<\/p>\n C\u00e2u 11:<\/b>\u00a0\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n <\/p>\n c\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn?<\/p>\n A. 1 \u00a0\u00a0\u00a0 B. 2 \u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0 D. 4<\/p>\n C\u00e2u 12:<\/b>\u00a0\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n <\/p>\n c\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn?<\/p>\n A. 1 \u00a0\u00a0\u00a0 B. 2\u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4<\/p>\n C\u00e2u 13:<\/b>\u00a0T\u00ecm m \u0111\u1ec3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n <\/p>\n c\u00f3 ba \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn<\/p>\n <\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 7:<\/b><\/p>\n Ta c\u00f3<\/p>\n <\/p>\n => y= -1 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 .<\/p>\n C\u00e2u 8:<\/b><\/p>\n Ta c\u00f3<\/p>\n <\/p>\n Do \u0111\u00f3 x – 2 = 0 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.<\/p>\n C\u00e2u 9:<\/b><\/p>\n y = m l\u00e0 ti\u1ec7m c\u1eadn ngang, x = m l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. V\u1eady giao \u0111i\u1ec3m hai ti\u1ec7m c\u1eadn l\u00e0 I(m;m). D\u1ec5 d\u00e0ng nh\u1eadn th\u1ea5y I thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh y=x.<\/p>\n C\u00e2u 11:<\/b><\/p>\n x= -3; y=1; y= -1 l\u00e0 ba \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.<\/p>\n C\u00e2u 13:<\/b><\/p>\n \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 ba \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ch\u1ec9 khi ph\u01b0\u01a1ng tr\u00ecnh x2<\/sup>\u00a0-2mx + 4 = 0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t kh\u00e1c -1.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 7:\u00a0\u0110\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 A.y = 1 \u00a0\u00a0\u00a0 B. y = 0 \u00a0\u00a0\u00a0 C. y = -1 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i C\u00e2u 8:\u00a0\u0110\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 A. x = 0 \u00a0\u00a0\u00a0 B. x = 2, x = -2 \u00a0\u00a0\u00a0 C. x – 2 […]<\/p>\n","protected":false},"author":3,"featured_media":27921,"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 7-C<\/td>\n 8-C<\/td>\n 9-A<\/td>\n 10-C<\/td>\n 11-D<\/td>\n 12-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n