1. H\u00e0m s\u1ed1 b\u1eadc ba y = f(x) = ax3<\/sup>\u00a0+ bx2<\/sup>\u00a0+ cx + d (a \u2260 0)<\/b><\/p>\n Ch\u00fa \u00fd:<\/b><\/p>\n – \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1eadc ba nh\u1eadn \u0111i\u1ec3m U(x0<\/sub>, y0<\/sub>) v\u1edbi x0<\/sub>\u00a0l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh f”(x) = 0 l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n – \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1eadc ba ho\u1eb7c c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb ho\u1eb7c kh\u00f4ng c\u00f3 \u0111i\u1ec3m c\u1ef1c tr\u1ecb n\u00e0o.<\/p>\n – \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1eadc ba lu\u00f4n c\u1eaft tr\u1ee5c ho\u00e0nh t\u1ea1i \u00edt nh\u1ea5t m\u1ed9t \u0111i\u1ec3m.<\/p>\n <\/p>\n 2. H\u00e0m s\u1ed1 b\u1eadc b\u1ed1n tr\u00f9ng ph\u01b0\u01a1ng:<\/b><\/p>\n <\/p>\n <\/p>\n Ch\u00fa \u00fd:<\/b><\/p>\n – \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax4<\/sup>\u00a0+ bx2<\/sup>\u00a0+ c c\u00f3 d\u1ea1ng (1) ho\u1eb7c (2) khi ab > 0 (a,b c\u00f9ng d\u1ea5u).<\/p>\n – \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax4<\/sup>\u00a0+ bx2<\/sup>\u00a0+ c c\u00f3 d\u1ea1ng (3) ho\u1eb7c (4) khi ab < 0 (a,b tr\u00e1i d\u1ea5u).<\/p>\n <\/p>\n <\/p>\n Ch\u00fa \u00fd:<\/b><\/p>\n – \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 hai \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn:<\/p>\n ti\u1ec7m c\u1eadn \u0111\u1ee9ng: x = -d\/c, ti\u1ec7m c\u1eadn ngang: y = a\/c<\/p>\n – \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n <\/p>\n nh\u1eadn giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn I(-d\/c; a\/c) l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n","protected":false},"excerpt":{"rendered":" 1. H\u00e0m s\u1ed1 b\u1eadc ba y = f(x) = ax3\u00a0+ bx2\u00a0+ cx + d (a \u2260 0) Ch\u00fa \u00fd: – \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1eadc ba nh\u1eadn \u0111i\u1ec3m U(x0, y0) v\u1edbi x0\u00a0l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh f”(x) = 0 l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng. – \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1eadc ba ho\u1eb7c c\u00f3 hai \u0111i\u1ec3m c\u1ef1c […]<\/p>\n","protected":false},"author":3,"featured_media":27913,"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