Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn<\/p>\n\n\n\n
To\u00e1n h\u1ecdc l\u00e0 m\u00f4n h\u1ecdc kh\u00f4ng ch\u1ec9 \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng t\u1ea1i tr\u01b0\u1eddng h\u1ecdc m\u00e0 n\u00f3 c\u00f2n \u0111\u01b0\u1ee3c \u00e1p d\u1ee5ng trong cu\u1ed9c s\u1ed1ng r\u1ea5t nhi\u1ec1u.V\u00e0 \u0111\u1ec3 h\u1ecdc t\u1ed1t to\u00e1n h\u1ecdc y\u00eau c\u1ea7u h\u1ecdc sinh<\/a> ph\u1ea3i si\u00eang n\u0103ng h\u1ecdc t\u1eadp<\/a>. H\u00f4m nay, Vietlearn s\u1ebd gi\u00fap cho c\u00e1c b\u1ea1n c\u00f3 th\u1ec3 ph\u00e2n t\u00edch v\u00e0 gi\u1ea3i m\u1ed9t b\u00e0i t\u1eadp c\u00f3 d\u1ea1ng quy \u0111\u1ed3ng m\u1eabu th\u1ee9c nhi\u1ec1u ph\u00e2n th\u1ee9c. H\u00e3y theo d\u00f5i ngay b\u00e0i vi\u1ebft \u0111\u1ec3 c\u00f3 th\u1ec3 hi\u1ec3u h\u01a1n v\u1ec1 d\u1ea1ng to\u00e1n n\u00e0y nh\u00e9!<\/p>\n\n\n\n L\u00fd thuy\u1ebft quy \u0111\u1ed3ng m\u1eabu th\u1ee9c nhi\u1ec1u ph\u00e2n th\u1ee9c<\/p>\n\n\n\n C\u00e1c t\u00ednh ch\u1ea5t quan tr\u1ecdng c\u1ee7a ph\u00e2n th\u1ee9c<\/p>\n\n\n\n \u0110\u1ec3 c\u00f3 th\u1ec3 gi\u1ea3i \u0111\u01b0\u1ee3c c\u00e1c b\u00e0i t\u1eadp v\u1ec1 quy \u0111\u1ed3ng m\u1eabu th\u1ee9c nhi\u1ec1u ph\u00e2n th\u1ee9c, \u0111i\u1ec1u ti\u00ean quy\u1ebft l\u00e0 c\u00e1c b\u1ea1n c\u1ea7n ph\u1ea3i n\u1eb1m l\u00f2ng c\u00e1c t\u00ednh ch\u1ea5t c\u01a1 b\u1ea3n c\u1ee7a ph\u00e2n th\u1ee9c. D\u00f9 l\u00e0 d\u1ea1ng to\u00e1n c\u01a1 b\u1ea3n hay n\u00e2ng cao th\u00ec ch\u1eafc ch\u1eafn ch\u00fang \u0111\u1ec1u \u00e1p d\u1ee5ng nh\u1eefng t\u00ednh ch\u1ea5t quan tr\u1ecdng n\u00e0y.<\/p>\n\n\n\n Cho ph\u00e2n th\u1ee9c A\/B v\u00e0 hai \u0111a th\u1ee9c M,N v\u1edbi \u0111i\u1ec1u ki\u1ec7n \u0111a th\u1ee9c M,N kh\u00e1c 0. Ta c\u00f3 c\u00e1c t\u00ednh ch\u1ea5t sau:<\/p>\n\n\n\n A\/B = (A.M) \/ (B.M)<\/p>\n\n\n\n A\/B = (A\/N) \/ (B\/N)<\/p>\n\n\n\n L\u01b0u \u00fd m\u1ed9t s\u1ed1 quy t\u1eafc \u0111\u1ed5i d\u1ea5u c\u1ee7a ph\u00e2n th\u1ee9c nh\u01b0 sau:<\/p>\n\n\n\n A\/B = -(-A)\/B<\/p>\n\n\n\n A\/B = -(A)\/-B<\/p>\n\n\n\n A\/(-B) = -(A)\/B<\/p>\n\n\n\n A\/B = -A\/-B<\/p>\n\n\n\n R\u00fat g\u1ecdn ph\u00e2n th\u1ee9c<\/p><\/blockquote>\n\n\n\n M\u1ed9t s\u1ed1 d\u1ea1ng b\u00e0i t\u1eadp quy \u0111\u1ed3ng m\u1eabu th\u1ee9c nhi\u1ec1u ph\u00e2n th\u1ee9c<\/p>\n\n\n\n D\u1ea1ng b\u00e0i quy \u0111\u1ed3ng ph\u00e2n th\u1ee9c \u0111\u01b0\u1ee3c \u0111\u00e1nh gi\u00e1 l\u00e0 kh\u00e1 \u0111\u01a1n gi\u1ea3n nh\u01b0ng c\u1ea7n kh\u1ea3 n\u0103ng ph\u00e2n t\u00edch t\u1ed1t \u0111\u1ec3 c\u00f3 th\u1ec3 t\u00ecm ra c\u00e1ch gi\u1ea3i nhanh nh\u1ea5t. Quy t\u1eafc chung th\u01b0\u1eddng \u0111\u01b0\u1ee3c \u00e1p d\u1ee5ng l\u00e0 t\u00ecm c\u00e1ch bi\u1ebfn \u0111\u1ed5i nh\u1eefng ph\u00e2n th\u1ee9c ph\u1ee9c t\u1ea1p th\u00e0nh nh\u1eefng ph\u00e2n th\u1ee9c c\u00f3 chung m\u1eabu s\u1ed1. Sau \u0111\u00f3 t\u00ednh to\u00e1n l\u1ea1i d\u1ea1ng r\u00fat g\u1ecdn c\u1ed9ng tr\u1eeb nh\u00e2n chia.<\/p>\n\n\n\n V\u1eady \u0111\u1ec3 c\u00f3 th\u1ec3 t\u00ednh to\u00e1n nh\u1eefng d\u1ea1ng b\u00e0i t\u1eadp quy \u0111\u1ed3ng m\u1eabu nhi\u1ec1u ph\u00e2n th\u1ee9c m\u1ed9t c\u00e1ch nhanh ch\u00f3ng m\u00e0 hi\u1ec7u qu\u1ea3, ch\u00fang ta c\u00f9ng nhau \u0111i t\u00ecm hi\u1ec3u v\u1ec1 c\u00e1ch th\u1ef1c hi\u1ec7n m\u1ed9t s\u1ed1 b\u00e0i t\u1eadp c\u1ee5 th\u1ec3 nh\u00e9!<\/p>\n\n\n\n B\u00e0i t\u1eadp quy \u0111\u1ed3ng m\u1eabu th\u1ee9c nhi\u1ec1u ph\u00e2n th\u1ee9c \u2013 T\u00ecm m\u1eabu th\u1ee9c chung<\/p>\n\n\n\n B\u01b0\u1edbc \u0111\u1ea7u ti\u00ean \u0111\u1ec3 c\u00f3 th\u1ec3 l\u00e0m \u0111\u01b0\u1ee3c c\u00e1c b\u00e0i t\u1eadp quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c th\u00ec ch\u00fang ta c\u1ea7n ph\u1ea3i l\u00e0m thao t\u00e1c \u0111i t\u00ecm m\u1eabu th\u1ee9c chung v\u00e0 \u0111\u1ec3 c\u00f3 th\u1ec3 t\u00ecm \u0111\u01b0\u1ee3c m\u1eabu th\u1ee9c chung ch\u00fang ta c\u00f3 th\u1ec3 l\u00e0m theo c\u00e1c b\u01b0\u1edbc sau:<\/p>\n\n\n\n Ph\u00e2n t\u00edch m\u1eabu th\u1ee9c c\u1ee7a c\u00e1c ph\u00e2n th\u1ee9c m\u00e0 cho tr\u01b0\u1edbc th\u00e0nh nh\u00e2n t\u1eed.<\/p>\n\n\n\n Nh\u00e2n t\u1eed chung b\u1eb1ng s\u1ed1 s\u1ebd b\u1eb1ng t\u00edch c\u1ee7a c\u00e1c nh\u00e2n t\u1eed b\u1eb1ng s\u1ed1 c\u1ee7a ph\u00e2n th\u1ee9c \u0111\u00e3 ph\u00e2n t\u00edch tr\u01b0\u1edbc \u0111\u00f3.<\/p>\n\n\n\n N\u1ebfu c\u00e1c nh\u00e2n t\u1eed b\u1eb1ng s\u1ed1 \u1edf m\u1eabu th\u1ee9c l\u00e0 nh\u1eefng s\u1ed1 nguy\u00ean kh\u00f4ng \u00e2m th\u00ec nh\u00e2n t\u1eed b\u1eb1ng s\u1ed1 s\u1ebd l\u00e0 b\u1ed9i chung nh\u1ecf nh\u1ea5t c\u1ee7a m\u1eabu th\u1ee9c.<\/p>\n\n\n\n V\u1edbi m\u1ed7i c\u01a1 s\u1ed1 c\u1ee7a l\u0169y th\u1eeba trong c\u00e1c m\u1eabu th\u1ee9c, ch\u00fang ta s\u1ebd ti\u1ebfn h\u00e0nh l\u1ef1a ch\u1ecdn l\u0169y th\u1eeba c\u00f3 s\u1ed1 m\u0169 cao nh\u1ea5t.<\/p>\n\n\n\n Nh\u1eefng d\u1ea1ng b\u00e0i t\u1eadp n\u00e0y s\u1ebd gi\u00fap b\u1ea1n ph\u00e1t tri\u1ec3n t\u01b0 duy t\u1ed1t h\u01a1n<\/p>\n\n\n\n V\u00ed d\u1ee5 1: T\u00ecm m\u1eabu th\u1ee9c chung c\u1ee7a hai th\u1ee9c 1\/(2×2 \u2013 4x + 2) v\u00e0 2\/(3x \u2013 3)<\/p>\n\n\n\n C\u00e1ch gi\u1ea3i:<\/p>\n\n\n\n Ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh c\u00e1c nh\u00e2n t\u1eed:<\/p>\n\n\n\n 2×2 \u2013 4x + 2 = 2(x2 \u2013 2x + 1) = 2(x \u2013 1)2<\/p>\n\n\n\n 3x \u2013 3 = 3(x \u2013 1)<\/p>\n\n\n\n M\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c b\u1eb1ng 6(x \u2013 1)2<\/p>\n\n\n\n X\u00e1c \u0111\u1ecbnh m\u1eabu th\u1ee9c chung nh\u1ecf nh\u1ea5t c\u1ee7a nguy\u00ean l\u00e0 b\u1ed9i chung nh\u1ecf nh\u1ea5t(2,3) = 6<\/p>\n\n\n\n M\u1eabu th\u1ee9c chung c\u1ee7a l\u0169y th\u1eeba (x \u2013 1) l\u00e0 (x \u2013 1)2<\/p>\n\n\n\n Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c<\/p>\n\n\n\n Sau khi th\u1ef1c hi\u1ec7n c\u00e1c b\u01b0\u1edbc t\u00ecm nh\u00e2n t\u1eed chung, ch\u00fang ta s\u1ebd ti\u1ebfn h\u00e0nh c\u00e1c b\u01b0\u1edbc quy \u0111\u1ed3ng ph\u00e2n th\u1ee9c:<\/p>\n\n\n\n Ph\u00e2n t\u00edch m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i t\u00ecm m\u1eabu th\u1ee9c chung.<\/p>\n\n\n\n T\u00ecm ra nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1ed7i m\u1eabu th\u1ee9c \u0111\u00e3 cho<\/p>\n\n\n\n L\u1ea5y t\u00edch t\u1eed v\u00e0 m\u1eabu c\u1ee7a m\u00f4i ph\u00e2n th\u1ee9c v\u1edbi nh\u00e2n t\u1eed ph\u1ee5 \u0111\u00e3 t\u00ecm \u0111\u01b0\u1ee3c.<\/p>\n\n\n\n V\u00ed d\u1ee5 2: Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u1ee7a hai ph\u00e2n th\u1ee9c 1\/(2×2 \u2013 4x + 2) v\u00e0 2\/(3x \u2013 3).<\/p>\n\n\n\n \u1ede v\u00ed d\u1ee5 1 ch\u00fang ta \u0111\u00e3 ph\u00e2n t\u00edch nh\u00e2n t\u1eed v\u00e0 t\u00ecm \u0111\u01b0\u1ee3c m\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c l\u00e0 6(x \u2013 1)2<\/p>\n\n\n\n T\u00ecm nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a hai m\u1eabu th\u1ee9c 2×2 \u2013 4x + 2 v\u00e0 3x \u2013 3<\/p>\n\n\n\n V\u00ec 6(x \u2013 1)2 = 3.2(x2 \u2013 2x + 1) = 3.(2×2 \u2013 4x + 2) n\u00ean 3 ch\u00ednh l\u00e0 nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c 2×2 \u2013 4x + 2<\/p>\n\n\n\n V\u00ec 6(x \u2013 1)2 = 2(x \u2013 1).3(x \u2013 1) n\u00ean 2(x \u2013 1) ch\u00ednh l\u00e0 nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c 3x \u2013 3<\/p>\n\n\n\n B\u00e0i t\u1eadp t\u1eadp v\u1eadn d\u1ee5ng quy \u0111\u1ed3ng m\u1eabu th\u1ee9c nhi\u1ec1u ph\u00e2n th\u1ee9c<\/p>\n\n\n\n Nh\u1eefng d\u1ea1ng b\u00e0i t\u1eadp th\u01b0\u1eddng g\u1eb7p<\/p>\n\n\n\n B\u00e0i 1: Cho hai ph\u00e2n th\u1ee9c:<\/p>\n\n\n\n T\u00ecm nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c x2 \u2013 4x2y + 4xy2 \u2013 y2<\/p>\n\n\n\n A. x(x \u2013 y).(x + y \u2013 4xy) B. x<\/p>\n\n\n\n C. x \u2013 y D. \u0110\u00e1p \u00e1n kh\u00e1c<\/p>\n\n\n\n B\u00e0i 2: T\u00ecm m\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c sau:<\/p>\n\n\n\n A. xy2(x + y) B. xy2<\/p>\n\n\n\n C. xy2(x + 1) D. xy(x + y + 1)<\/p>\n\n\n\n B\u00e0i 3: Hai ph\u00e2n th\u1ee9c (x + 1)\/(x2 + 2x \u2013 3) v\u00e0 (- 2x)\/(x2 + 7x + 10) c\u00f3 m\u1eabu th\u1ee9c chung l\u00e0?<\/p>\n\n\n\n x3 + 6×2 + 3x \u2013 10<\/p>\n\n\n\n x3 \u2013 6×2 + 3x \u2013 10<\/p>\n\n\n\n x3 + 6×2 \u2013 3x \u2013 10<\/p>\n\n\n\n \u0110\u00e1p \u00e1n kh\u00e1c<\/p>\n\n\n\n B\u00e0i 2: Hai ph\u00e2n th\u1ee9c 5\/(2x + 6) v\u00e0 3\/(x2 \u2013 9) c\u00f3 m\u1eabu th\u1ee9c chung \u0111\u01a1n gi\u1ea3n nh\u1ea5t l\u00e0 ?<\/p>\n\n\n\n x2 \u2013 9.<\/p>\n\n\n\n 2( x2 \u2013 9 ).<\/p>\n\n\n\n x2 + 9.<\/p>\n\n\n\n x \u2013 3<\/p>\n\n\n\n L\u1eddi k\u1ebft<\/p>\n\n\n\n Tr\u00ean \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 ki\u1ebfn th\u1ee9c c\u01a1 b\u1ea3n<\/a> v\u1ec1 quy \u0111\u1ed3ng m\u1eabu th\u1ee9c nhi\u1ec1u ph\u00e2n th\u1ee9c m\u00e0 Vietlearn mu\u1ed1n gi\u1edbi thi\u1ec7u \u0111\u1ebfn cho c\u00e1c b\u1ea1n h\u1ecdc sinh<\/a>, v\u1edbi mong mu\u1ed1n gi\u00fap c\u00e1c b\u1ea1n d\u1ec5 d\u00e0ng ti\u1ebfp thu d\u1ea1ng b\u00e0i t\u1eadp to\u00e1n 8 quy \u0111\u1ed3ng m\u1eabu th\u1ee9c nhi\u1ec1u ph\u00e2n th\u1ee9c n\u00e0y h\u01a1n. N\u1ebfu c\u00e1c b\u1ea1n c\u00f2n mu\u1ed1n t\u00ecm hi\u1ec3u th\u00eam b\u1ea5t k\u1ef3 b\u00e0i h\u1ecdc n\u00e0o, \u0111\u1eebng ng\u1ea7n ng\u1ea1i truy c\u1eadp ngay v\u00e0o website: Vietlearn.org\/ \u0111\u1ec3 c\u00f3 nh\u1eefng gi\u00e2y ph\u00fat h\u1ecdc t\u1eadp<\/a> th\u00fa v\u1ecb nh\u00e9! Ch\u00fac c\u00e1c b\u1ea1n \u0111\u1ea1t \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 h\u1ecdc t\u1eadp<\/a> th\u1eadt t\u1ed1t!<\/p>\n\n\n\n T\u00ecm hi\u1ec3u th\u00eam:<\/p>\n\n\n\n Ph\u00e9p c\u1ed9ng c\u00e1c ph\u00e2n th\u1ee9c \u0111\u1ea1i s\u1ed1 \u2013 B\u00ed quy\u1ebft c\u1ea3i thi\u1ec7n m\u00f4n to\u00e1n 8<\/p>\n\n\n\n\n\n\n","protected":false},"excerpt":{"rendered":" Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn To\u00e1n h\u1ecdc l\u00e0 m\u00f4n h\u1ecdc kh\u00f4ng ch\u1ec9 \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng t\u1ea1i tr\u01b0\u1eddng h\u1ecdc m\u00e0 n\u00f3 c\u00f2n \u0111\u01b0\u1ee3c \u00e1p d\u1ee5ng trong cu\u1ed9c s\u1ed1ng r\u1ea5t nhi\u1ec1u.V\u00e0 \u0111\u1ec3 h\u1ecdc t\u1ed1t to\u00e1n h\u1ecdc y\u00eau c\u1ea7u h\u1ecdc sinh ph\u1ea3i si\u00eang n\u0103ng h\u1ecdc t\u1eadp. H\u00f4m nay, Vietlearn s\u1ebd gi\u00fap cho c\u00e1c b\u1ea1n c\u00f3 th\u1ec3 ph\u00e2n […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":"","_disable_autopaging":false},"categories":[9,4],"tags":[],"class_list":["post-1073","post","type-post","status-publish","format-standard","hentry","category-toan-hoc","category-trung-hoc"],"_links":{"self":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/1073","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/comments?post=1073"}],"version-history":[{"count":0,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/1073\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/media?parent=1073"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/categories?post=1073"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/tags?post=1073"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}