D\u01b0\u1edbi \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 d\u1ea1ng to\u00e1n chuy\u00ean \u0111\u1ec1 v\u1ec1 c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 m\u00e0 ch\u00fang ta th\u01b0\u1eddng g\u1eb7p trong c\u00e1c \u0111\u1ec1 th\u00ec. Gi\u00fap b\u1ea1n l\u00e0m quen v\u1edbi c\u00e1ch gi\u1ea3i nhanh ch\u00f3ng v\u00e0 ch\u00ednh x\u00e1c.<\/p>\n

1\/ D\u1ea1ng 1: Chuy\u00ean \u0111\u1ec1 c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 c\u00f3 d\u1ea1ng t\u00ecm m \u0111\u1ec3 h\u00e0m s\u1ed1\u00a0y = f(x)\u00a0\u0111\u1ea1t c\u1ef1c \u0111\u1ea1i ho\u1eb7c c\u1ef1c ti\u1ec3u t\u1ea1i\u00a0<\/strong>x0<\/sub><\/h2>\n

Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i<\/strong>: ta s\u1eed d\u1ee5ng \u0111i\u1ec1u ki\u1ec7n sau \u0111\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n d\u1ea1ng n\u00e0y:<\/p>\n

\n
• N\u1ebfu\u00a0\u00a0th\u00ec h\u00e0m s\u1ed1 f(x)\u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i\u00a0\u0111i\u1ec3m x_{0<\/sub>\u00a0.<\/li>\n}
• N\u1ebfu\u00a0\u00a0th\u00ec h\u00e0m s\u1ed1 f(x) \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i\u00a0x0<\/sub>.<\/li>\n<\/ul>\n

  V\u00ed d\u1ee5 <\/strong>: T\u00ecm\u00a0m \u0111\u1ec3 h\u00e0m s\u1ed1\u00a0\u00a0\u00a0y = 1\/3 x^{3<\/sup> + (m2 <\/sup>\u2013 m + 2)x2<\/sup> + (3m2<\/sup> + 1)x \u00a0+ m -5 \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c ti\u1ec3u t\u1ea1i\u00a0x<\/em>\u00a0= -2.<\/p>\n}

  Gi\u1ea3i:<\/strong><\/p>\n

  Ta c\u00f3: y’ (x) = x\u00b2 + 2(m\u00b2 – m + 2)x +3m\u00b2 +1<\/p>\n

  \u21d2y”(x) = 2x + 2(m\u00b2 – m +2)<\/p>\n

  \u0110\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = -2 th\u00ec y'(-2) = 0<\/p>\n

  \u21d4 – m\u00b2 + 4m – 3 = 0\u00a0\u21d4 m = 1 v\u00e0 m=3<\/p>\n

  V\u1edbi m = 3 ta c\u00f3 y(-2) = 8 > 0 n\u00ean h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = -2<\/p>\n

  V\u1edbi m = 1 ta c\u00f3 y(-2) = (1 \/ 3)x\u00b3 + 2x\u00b2 + 4x – 4 x\u00e9t b\u1ea3ng bi\u1ebfn thi\u00ean th\u00ec ta th\u1ea5y h\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb t\u1ea1i x = -2 n\u00ean m =1 kh\u00f4ng th\u1ecfa m\u00e3n y\u00eau c\u1ea7u<\/p>\n

  V\u1eady: v\u1edbi m = 3 th\u00ec h\u00e0m s\u1ed1 \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c ti\u1ec3u t\u1ea1i x = -2<\/p>\n

  2\/ D\u1ea1ng 2:<\/h2>\n

  D\u1ea1ng 2: T\u00ecm m \u0111\u1ec3 h\u00e0m s\u1ed1\u00a0y = f(x)\u00a0c\u00f3 c\u1ef1c tr\u1ecb ho\u1eb7c kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb.<\/strong><\/p>\n

  \u0110\u1ed1i v\u1edbi d\u1ea1ng b\u00e0i to\u00e1n n\u00e0y ch\u00fang ta th\u01b0\u1eddng ch\u00fa \u00fd \u0111\u1ebfn hai d\u1ea1ng h\u00e0m s\u1ed1 ch\u00ednh l\u00e0:<\/p>\n

  a\/ Chuy\u00ean \u0111\u1ec1 c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 b\u1eadc 3:\u00a0y = ax3 <\/sup>+ bx2<\/sup> + cx + d (v\u1edbi a # 0)<\/h3>\n

  H\u00e0m s\u1ed1 y kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb\u00a0\u00f3\u00a0ph\u01b0\u01a1ng tr\u00ecnh\u00a0y\u2019 = 0 l\u00e0 v\u00f4 nghi\u1ec7m ho\u1eb7c c\u00f3 nghi\u1ec7m k\u00e9p\u00a0\u00f3 \u2206 < 0\u00a0.<\/p>\n

  H\u00e0m s\u1ed1 c\u00f3 2 \u0111i\u1ec3m \u00a0c\u1ef1c tr\u1ecb\u00a0\u00f3 ph\u01b0\u01a1ng tr\u00ecnh y\u2019 = 0\u00a0c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t\u00a0\u00f3 \u2206 > 0<\/p>\n

  b\/ Chuy\u00ean \u0111\u1ec1 c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 b\u1eadc 4 tr\u00f9ng ph\u01b0\u01a1ng:\u00a0 y = ax4<\/sup> + bx2<\/sup> + c (v\u1edbi a # 0)<\/h3>\n

  <\/p>\n

  H\u00e0m s\u1ed1 c\u00f3 1 \u0111i\u1ec3m c\u1ef1c tr\u1ecb\u00a0\u00f3\u00a0ph\u01b0\u01a1ng tr\u00ecnh\u00a0y\u2019 = 0\u00a0c\u00f3 1 nghi\u1ec7m duy nh\u1ea5t\u00a0\u00f3\u00a0\u00a0a.b > 0.<\/p>\n

  H\u00e0m s\u1ed1 c\u00f3 3 c\u1ef1c tr\u1ecb\u00a0\u00f3\u00a0ph\u01b0\u01a1ng tr\u00ecnh\u00a0y\u2019 = 0\u00a0c\u00f3 3 nghi\u1ec7m\u00a0\u00f3\u00a0\u00a0a.b < 0.<\/p>\n

  V\u00ed d\u1ee5 <\/strong>: Cho h\u00e0m s\u1ed1\u00a0sau : y = f(x) = mx^{3<\/sup> + 3mx2 <\/sup>\u2013 (m-1)x – 1, v\u1edbi m l\u00e0 tham s\u1ed1. X\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb c\u1ee7a m sao cho h\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb.<\/p>\n}

  Gi\u1ea3i<\/u><\/strong><\/p>\n

  V\u1edbi m = 0 ta c\u00f3 y = x \u2013 1 \u00a0n\u00ean h\u00e0m s\u1ed1 y = f(x) s\u1ebd kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb.<\/p>\n

  V\u1edbi m # 0 ta c\u00f3 y\u2019 = 3mx^{2 <\/sup>+ 6mx \u2013( m \u2013 1)<\/p>\n}

  H\u00e0m s\u1ed1 y = f(x) \u00a0kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb \u00f3ph\u01b0\u01a1ng tr\u00ecnh y’ = 0 v\u00f4 nghi\u1ec7m ho\u1eb7c l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p.<\/p>\n

  \u2206\u2019 = 9m^{2<\/sup> + 3m(m – 1) = 12m2<\/sup> \u2013 3m < 0 \u00f3 0<=m<=1\/4<\/p>\n}

  V\u1eady v\u1edbi\u00a00<=m<=1\/4\u00a0th\u00ec h\u00e0m s\u1ed1 y = f(x) kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb.<\/p>\n

  3\/ D\u1ea1ng 3: T\u00ecm tham s\u1ed1 m \u0111\u1ec3 h\u00e0m s\u1ed1 <\/strong>y = f(x) c\u00f3 c\u1ef1c tr\u1ecb th\u1ecfa m\u00e3n y\u00eau c\u1ea7u.<\/strong><\/h2>\n

  \u0110\u00e2y l\u00e0 1 d\u1ea1ng b\u00e0i t\u1eadp n\u00e2ng cao ch\u00fang ta th\u01b0\u1eddng g\u1eb7p trong c\u00e1c \u0111\u1ec1 thi \u0111\u1ea1i h\u1ecdc v\u00e0 cao \u0111\u1eb3ng. \u0110\u1ec3 c\u00f3 th\u1ec3 gi\u1ea3i \u0111\u01b0\u1ee3c d\u1ea1ng to\u00e1n n\u00e0y tr\u01b0\u1edbc ti\u00ean ch\u00fang \u00a0ta c\u1ea7n n\u1eafm \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng ph\u00e1p \u0111\u1ec3 gi\u1ea3i c\u00e1c d\u1ea1ng to\u00e1n b\u00ean tr\u00ean \u0111\u1ed3ng th\u1eddi k\u1ebft h\u1ee3p v\u1edbi 1 s\u1ed1 ki\u1ebfn th\u1ee9c kh\u00e1c v\u1ec1 h\u00ecnh h\u1ecdc, d\u00e3y s\u1ed1… \u0111\u1ec3 c\u00f3 th\u1ec3 gi\u1ea3i d\u1ea1ng to\u00e1n n\u00e0y<\/p>\n

  V\u00ed d\u1ee5 <\/strong>: Cho h\u00e0m s\u1ed1\u00a0y = f(x) = x^{4<\/sup> + 2m2<\/sup>x2<\/sup> + 1 (C_{m<\/sub>). T\u00ecm m \u0111\u1ec3 h\u00e0m s\u1ed1 y = f(x) c\u00f3 3 \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0 3 \u0111\u1ec9nh c\u1ee7a 1 tam gi\u00e1c vu\u00f4ng c\u00e2n.<\/p>\n}}

  Gi\u1ea3i<\/u><\/strong><\/p>\n

  Tr\u01b0\u1edbc ti\u00ean ch\u00fang ta \u00e1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i \u1edf d\u1ea1ng 2 \u0111\u1ec3 t\u00ecm m sao cho h\u00e0m s\u1ed1 y = f(x) \u00a0c\u00f3 3 c\u1ef1c tr\u1ecb.<\/p>\n

  Ta c\u00f3:\u00a0y\u2019 = 4x^{3<\/sup> \u2013 4m2<\/sup>x = 4x(x2<\/sup> \u2013 m2<\/sup>)<\/p>\n}

  V\u1edbi y\u2019 = 0 ta c\u00f3 y\u2019 = 0 \u00f3 4x(x^{2<\/sup> \u2013 m2<\/sup>) = 0 \u00f3 x = 0 v\u00e0 x2 <\/sup>= m2<\/sup><\/p>\n}

  \u0110\u1ec3 h\u00e0m s\u1ed1 y = f(x) c\u00f3 3 c\u1ef1c tr\u1ecb th\u00ec ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh y’ = 0 ph\u1ea3i c\u00f3 3 nghi\u1ec7m ph\u00e2n bi\u1ec7t.<\/p>\n

  \u00f3\u00a0Ph\u01b0\u01a1ng tr\u00ecnh x^{2 <\/sup>= m2<\/sup> ph\u1ea3i c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t v\u00e0 kh\u00e1c\u00a00 \u00f3 m # 0<\/p>\n}

  V\u1eady v\u1edbi\u00a0m # 0\u00a0th\u00ec h\u00e0m s\u1ed1 c c\u00f3 3 c\u1ef1c tr\u1ecb.<\/p>\n

  Ta c\u1ea7n t\u00ecm m \u0111\u1ec3 3 c\u1ef1c tr\u1ecb n\u00e0y s\u1ebd t\u1ea1o th\u00e0nh 3 \u0111\u1ec9nh c\u1ee7a 1 tam gi\u00e1c vu\u00f4ng c\u00e2n.<\/p>\n

  Ta c\u00f3: v\u1edbi\u00a0m # 0\u00a0\u00a0th\u00ec y\u2019 = 0<\/p>\n

  V\u1edbi \u00a0x = 0 => y = 1<\/p>\n

  V\u1edbi x = m => y = 1 \u2013 m^{4<\/sup><\/p>\n}

  V\u1edbi x = – m => y = 1- m^{4<\/sup><\/p>\n}

  G\u1ecdi 3 \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = f(x) l\u1ea7n l\u01b0\u1ee3t l\u00e0: A (0 ; 1)\u00a0B (- m ; 1 \u2013 m^{4<\/sup>) C (m ; 1 \u2013 m4<\/sup>)<\/p>\n}

  Theo t\u00ednh ch\u1ea5t c\u1ee7a 1 h\u00e0m s\u1ed1 b\u1eadc 4 tr\u00f9ng ph\u01b0\u01a1ng th\u00ec ta c\u00f3 tam gi\u00e1c ABC c\u00e2n t\u1ea1i A n\u00ean \u0111\u1ec3 tam gi\u00e1c \u00a0ABC vu\u00f4ng c\u00e2n th\u00ec c\u1ea1nh \u00a0AB vu\u00f4ng g\u00f3c v\u1edbi c\u1ea1nh \u00a0AC<\/p>\n

  Ta c\u00f3 AB.AC = 0<\/p>\n

  <=> ta c\u00f3 m = 0 (lo\u1ea1i)<\/p>\n

  V\u1edbi m = – 1; m = 1 ( th\u1ecfa m\u00e3n)<\/p>\n

  V\u1eady v\u1edbi m = -1 v\u00e0 m = 1 th\u00ec th\u1ecfa m\u00e3n y\u00eau c\u1ea7u c\u1ee7a b\u00e0i to\u00e1n.<\/p>\n

  V\u00ed d\u1ee5:<\/strong><\/p>\n

  <\/p>\n

  Tr\u00ean \u0111\u00e2y l\u00e0 3 d\u1ea1ng to\u00e1n v\u1ec1 chuy\u00ean \u0111\u1ec1 c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 m\u00e0 ch\u00fang ta th\u01b0\u1eddng g\u1eb7p. Trong \u0111\u00f3 c\u00f3 d\u1ea1ng 1 v\u00e0 2 l\u00e0 c\u00e1c d\u1ea1ng to\u00e1n c\u01a1 b\u1ea3n ch\u00fang ta c\u1ea7n ph\u1ea3i n\u1eafm v\u1eefng tr\u01b0\u1edbc khi ch\u00fang ta t\u00ecm hi\u1ec3u \u0111\u1ebfn d\u1ea1ng 3.<\/p>\n

   <\/p>\n","protected":false},"excerpt":{"rendered":"

  D\u01b0\u1edbi \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 d\u1ea1ng to\u00e1n chuy\u00ean \u0111\u1ec1 v\u1ec1 c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 m\u00e0 ch\u00fang ta th\u01b0\u1eddng g\u1eb7p trong c\u00e1c \u0111\u1ec1 th\u00ec. Gi\u00fap b\u1ea1n l\u00e0m quen v\u1edbi c\u00e1ch… <\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[450,470],"tags":[],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts\/8512"}],"collection":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/comments?post=8512"}],"version-history":[{"count":0,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts\/8512\/revisions"}],"wp:attachment":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/media?parent=8512"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/categories?post=8512"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/tags?post=8512"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}