Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn<\/p>\n\n\n\n
Nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c l\u00e0 d\u1ea1ng to\u00e1n \u0111\u1ea7u ti\u00ean m\u00e0 c\u00e1c b\u1ea1n h\u1ecdc sinh<\/a> ph\u1ea3i chinh ph\u1ee5c n\u1ebfu nh\u01b0 mu\u1ed1n h\u1ecdc t\u1ed1t m\u00f4n to\u00e1n. Tuy l\u00e0 d\u1ea1ng to\u00e1n \u0111\u01a1n gi\u1ea3n nh\u01b0ng n\u1ebfu kh\u00f4ng t\u1eadp trung t\u00ednh to\u00e1n r\u1ea5t d\u1ec5 x\u1ea3y ra l\u1ed7i sai. V\u00ec v\u1eady h\u00f4m nay, Vietlearn s\u1ebd \u0111em \u0111\u1ebfn cho c\u00e1c b\u1ea1n nh\u1eefng ki\u1ebfn th\u1ee9c tr\u1ecdng t\u00e2m nh\u1ea5t v\u1ec1 to\u00e1n 8 nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c \u0111\u1ec3 m\u1ecdi ng\u01b0\u1eddi c\u00f3 th\u1ec3 n\u1eafm v\u1eefng.<\/p>\n\n\n\n L\u00fd thuy\u1ebft v\u00e0 b\u00e0i t\u1eadp nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c<\/p>\n\n\n\n L\u00fd thuy\u1ebft<\/p>\n\n\n\n Quy t\u1eafc nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c c\u1ea7n ph\u1ea3i nh\u1edb<\/p>\n\n\n\n Mu\u1ed1n th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c n\u00e2ng cao ta l\u1ea5y \u0111\u01a1n th\u1ee9c nh\u00e2n cho t\u1eebng h\u1ea1ng t\u1eed c\u1ee7a \u0111a th\u1ee9c r\u1ed3i sau \u0111\u00f3 c\u1ed9ng c\u00e1c t\u00edch l\u1ea1i v\u1edbi nhau.<\/p>\n\n\n\n C\u00f4ng th\u1ee9c t\u1ed5ng qu\u00e1t: Cho A, B, ,C, D l\u00e0 c\u00e1c \u0111\u01a1n th\u1ee9c ta c\u00f3<\/p>\n\n\n\n A(B + C) = AB + AC<\/p>\n\n\n\n A(B + C \u2013 D) = AB + AC \u2013 AD<\/p>\n\n\n\n V\u00ed d\u1ee5: x (x2 + 2)<\/p>\n\n\n\n C\u00f4ng th\u1ee9c c\u1ee7a c\u00e1c ph\u00e9p t\u00ednh v\u1ec1 l\u0169y th\u1eeba<\/p>\n\n\n\n M\u1ed9t s\u1ed1 c\u00f4ng th\u1ee9c l\u0169y th\u1eeba m\u00e0 c\u00e1c b\u1ea1n h\u1ecdc sinh<\/a> ph\u1ea3i nh\u1edb \u0111\u1ec3 c\u00f3 th\u1ec3 l\u00e0m \u0111\u01b0\u1ee3c b\u00e0i t\u1eadp d\u1ea1ng nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c to\u00e1n l\u1edbp 8:<\/p>\n\n\n\n an = a . a . a \u2026 a (a \u2208 Q, n \u2208 N*)<\/p>\n\n\n\n a0 = 1 (a \u22600)<\/p>\n\n\n\n an . am = an + m<\/p>\n\n\n\n an : am = an \u2013 m (n \u2265 m)<\/p>\n\n\n\n (am)n = am . n<\/p>\n\n\n\n C\u00e1c d\u1ea1ng b\u00e0i t\u1eadp c\u01a1 b\u1ea3n<\/p>\n\n\n\n C\u00f3 3 d\u1ea1ng b\u00e0i t\u1eadp c\u01a1 b\u1ea3n khi l\u1ea5y \u0111\u01a1n th\u1ee9c nh\u00e2n \u0111a th\u1ee9c<\/p>\n\n\n\n Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh hay r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<\/p>\n\n\n\n S\u1eed d\u1ee5ng quy t\u1eafc nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c to\u00e1n 8 l\u00e0 c\u00f3 th\u1ec3 d\u1ec5 d\u00e0ng l\u00e0m \u0111\u01b0\u1ee3c d\u1ea1ng b\u00e0i t\u1eadp n\u00e0y.<\/p>\n\n\n\n V\u00ed d\u1ee5 1:<\/p>\n\n\n\n A = x.(x2 \u2013 y) \u2013 x2.(x + y) + y.(x2 \u2013 x)<\/p>\n\n\n\n V\u00ed d\u1ee5 2:<\/p>\n\n\n\n B = x(x \u2013 y) + y(x + y)<\/p>\n\n\n\n = x.x \u2013 x.y + y.x + y.y<\/p>\n\n\n\n = x2 \u2013 xy + xy + y2<\/p>\n\n\n\n = x2 + y2.<\/p>\n\n\n\n T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c<\/p>\n\n\n\n T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a f(x) t\u1ea1i x0<\/p>\n\n\n\n V\u00ed d\u1ee5 1: T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x.(x2 \u2013 y) \u2013 x2.(x + y) + y.(x2 \u2013 x) t\u1ea1i x = 6, y = 5<\/p>\n\n\n\n A = x.(x2 \u2013 y) \u2013 x2.(x + y) + y.(x2 \u2013 x)<\/p>\n\n\n\n Thay x = 6, y = 5 v\u00e0o v\u00e0o bi\u1ec3u th\u1ee9c A = \u20132xy<\/p>\n\n\n\n => A = -2 * 6 *5<\/p>\n\n\n\n => A = -60<\/p>\n\n\n\n V\u00ed d\u1ee5 2: T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c B = x(x \u2013 y) + y(x + y) t\u1ea1i x = 1, y =-2<\/p>\n\n\n\n B = x(x \u2013 y) + y(x + y)<\/p>\n\n\n\n = x.x \u2013 x.y + y.x + y.y<\/p>\n\n\n\n = x2 \u2013 xy + xy + y2<\/p>\n\n\n\n = x2 + y2<\/p>\n\n\n\n Thay x = 1, y = 2 v\u00e0o bi\u1ec3u th\u1ee9c B = x2 + y2<\/p>\n\n\n\n => B = 12 + 22<\/p>\n\n\n\n => B = 5<\/p>\n\n\n\n T\u00ecm X<\/p>\n\n\n\n S\u1eed d\u1ee5ng c\u00e1c quy t\u1eafc nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i bi\u1ebfn x v\u1ec1 d\u1ea1ng c\u01a1 b\u1ea3n.<\/p>\n\n\n\n V\u00ed d\u1ee5 1: 36×2 \u2013 12x + 9x(4x \u2013 3) = 30. T\u00ecm x = ?<\/p>\n\n\n\n 3x(12x \u2013 4) \u2013 9x(4x \u2013 3) = 30<\/p>\n\n\n\n 3x.12x \u2013 3x.4 \u2013 (9x.4x \u2013 9x.3) = 30<\/p>\n\n\n\n 36×2 \u2013 12x \u2013 36×2 + 27x = 30<\/p>\n\n\n\n (36×2 \u2013 36×2) + (27x \u2013 12x) = 30<\/p>\n\n\n\n 15x = 30<\/p>\n\n\n\n x = 2<\/p>\n\n\n\n V\u1eady x = 2.<\/p>\n\n\n\n V\u00ed d\u1ee5 2: x(5 \u2013 2x) + 2x(x \u2013 1) = 15. T\u00ecm x = ?<\/p>\n\n\n\n (x.5 \u2013 x.2x) + (2x.x \u2013 2x.1) = 15<\/p>\n\n\n\n 5x \u2013 2×2 + 2×2 \u2013 2x = 15<\/p>\n\n\n\n (2×2 \u2013 2×2) + (5x \u2013 2x) = 15<\/p>\n\n\n\n 3x = 15<\/p>\n\n\n\n x = 5.<\/p>\n\n\n\n V\u1eady x = 5<\/p>\n\n\n\n Nh\u00e2n \u0111a th\u1ee9c v\u1edbi \u0111a th\u1ee9c<\/p><\/blockquote>\n\n\n\n L\u00fd do c\u1ea7n l\u00e0m c\u00e1c d\u1ea1ng nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c b\u00e0i t\u1eadp<\/p>\n\n\n\n Nh\u00e2n \u0111\u01a1n th\u1ee9c \u0111a th\u1ee9c \u0111\u01b0\u1ee3c xem l\u00e0 d\u1ea1ng b\u00e0i t\u1eadp \u0111\u01a1n gi\u1ea3n, kh\u00f4ng qu\u00e1 kh\u00f3 ch\u1ec9 c\u1ea7n h\u1ecdc sinh<\/a> t\u1eadp trung khi l\u00e0m th\u00ec k\u1ebft qu\u1ea3 s\u1ebd r\u1ea5t t\u1ed1t. \u0110\u00e2y l\u00e0 d\u1ea1ng b\u00eca t\u1eadp c\u01a1 b\u1ea3n gi\u00fap h\u1ecdc sinh<\/a> c\u00f3 th\u1ec3 h\u00ecnh dung v\u00e0 t\u01b0 duy \u0111\u1ec3 sau n\u00e0y c\u00f3 th\u1ec3 h\u1ecdc t\u1ed1t v\u00e0 l\u00e0m \u0111\u01b0\u1ee3c nh\u1eefng d\u1ea1ng b\u00e0i t\u1eadp kh\u00f3 h\u01a1n, y\u00eau c\u1ea7u kh\u1ea3 n\u0103ng t\u01b0 duy v\u00e0 ph\u00e2n t\u00edch v\u00f4 c\u00f9ng cao.<\/p>\n\n\n\n Tuy nhi\u00ean, r\u1ea5t nhi\u1ec1u b\u1ea1n h\u1ecdc sinh<\/a> ch\u1ee7 quan cho r\u1eb1ng nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c l\u1edbp 8 l\u00e0 nh\u1eefng c\u00e2u b\u00e0i t\u1eadp \u0111\u01a1n gi\u1ea3n, v\u00ec v\u1eady kh\u00f4ng t\u1eadp trung v\u00e0o vi\u1ec7c h\u1ecdc, ph\u00e2n t\u00edch c\u00e1ch gi\u1ea3i m\u00e0 ch\u1ec9 l\u00e0m qua loa \u0111\u1ec3 cho xong v\u00ec v\u1eady d\u1eabn \u0111\u1ebfn t\u00ednh tr\u1ea1ng k\u1ebft qu\u1ea3 h\u1ecdc t\u1eadp<\/a> ng\u00e0y c\u00e0ng sa s\u00fat. Nh\u01b0 v\u1eady, c\u00f3 th\u1ec3 th\u1ea5y c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp ch\u00ednh l\u00e0 ti\u1ec1n \u0111\u1ec1 \u0111\u1ec3 cho h\u1ecdc sinh<\/a> c\u00f3 th\u1ec3 ph\u00e1t tri\u1ec3n t\u01b0 duy v\u00e0 kh\u1ea3 n\u0103ng ph\u00e2n t\u00edch cho nh\u1eefng d\u1ea1ng b\u00eca t\u1eadp kh\u00f3 h\u01a1n.<\/p>\n\n\n\n B\u00e0i t\u1eadp th\u1ef1c h\u00e0nh<\/p>\n\n\n\n B\u00e0i t\u1eadp th\u1ef1c h\u00e0nh v\u00e0 l\u1eddi gi\u1ea3i chi ti\u1ebft<\/p>\n\n\n\n B\u00e0i 1. L\u00e0m t\u00ednh nh\u00e2n:<\/p>\n\n\n\n a) x2(5×3 \u2013 x \u2013 1\/2);<\/p>\n\n\n\n b) (3xy \u2013 x2 + y). 2\/3x2y;<\/p>\n\n\n\n c) (4×3\u2013 5xy + 2x)(-1\/2xy).<\/p>\n\n\n\n \u0110\u00e1p \u00e1n v\u00e0 h\u01b0\u1edbng d\u1eabn gi\u1ea3i chi ti\u1ebft:<\/p>\n\n\n\n a) x2(5×3 \u2013 x \u20131\/2) = x2. 5×3 + x2 . (-x) + x2 . (-1\/2)<\/p>\n\n\n\n = 5×5 \u2013 x3 \u2013 1\/2×2<\/p>\n\n\n\n b) (3xy \u2013 x2 + y).2\/3x2y = 2\/3x2y. 3xy +2\/3x2y. (- x2) + 2\/3x2y. y<\/p>\n\n\n\n = 2x3y2 \u2013 2\/3x4y + 2\/3x2y2<\/p>\n\n\n\n c) (4×3\u2013 5xy + 2x)(-1\/2xy) = -1\/2xy . 4×3 + (-1\/2xy) . (-5xy) + (- 1\/2xy) . 2x<\/p>\n\n\n\n = -2x4y +5\/2x2y2 \u2013 x2y.<\/p>\n\n\n\n B\u00e0i 2.<\/p>\n\n\n\n Th\u1ef1c hi\u1ec7n ph\u00e9p nh\u00e2n, r\u00fat g\u1ecdn r\u1ed3i t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n a) x(x \u2013 y) + y(x + y) t\u1ea1i x = -6 v\u00e0 y = 8;<\/p>\n\n\n\n b) x(x2 \u2013 y) \u2013 x2 (x + y) + y (x2 \u2013 x) t\u1ea1i x =1\/2 v\u00e0 y = -100.<\/p>\n\n\n\n \u0110\u00e1p \u00e1n v\u00e0 h\u01b0\u1edbng d\u1eabn gi\u1ea3i chi ti\u1ebft;<\/p>\n\n\n\n a) x(x \u2013 y) + y (x + y) = x2 \u2013 xy +yx + y2= x2+ y2<\/p>\n\n\n\n v\u1edbi x = -6, y = 8 bi\u1ec3u th\u1ee9c c\u00f3 gi\u00e1 tr\u1ecb l\u00e0 (-6)2 + 82 = 36 + 64 = 100<\/p>\n\n\n\n b) x(x2 \u2013 y) \u2013 x2 (x + y) + y (x2\u2013 x) = x3 \u2013 xy \u2013 x3 \u2013 x2y + yx2 \u2013 yx = \u2013 2xy<\/p>\n\n\n\n V\u1edbi x =1\/2, y = -100 bi\u1ec3u th\u1ee9c c\u00f3 gi\u00e1 tr\u1ecb l\u00e0 -2 . 1\/2. (-100) = 100.<\/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 v\u1ec1 c\u00e1ch t\u00ednh nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c m\u00e0 Vietlearn mu\u1ed1n g\u1eedi \u0111\u1ebfn cho c\u00e1c b\u1ea1n h\u1ecdc sinh v\u1edbi hy v\u1ecdng s\u1ebd gi\u00fap c\u00e1c b\u1ea1n d\u1ec5 d\u00e0ng ti\u1ebfp thu ki\u1ebfn th\u1ee9c v\u1ec1 c\u00e1ch gi\u1ea3i nh\u1eefng b\u00e0i to\u00e1n n\u00e0y h\u01a1n. N\u1ebfu c\u00e1c b\u1ea1n c\u00f2n mu\u1ed1n t\u00ecm hi\u1ec3u th\u00eam c\u00e1c c\u00e1ch t\u00ednh kh\u00e1c v\u1ec1 ki\u1ebfn th\u1ee9c m\u00f4n to\u00e1n th\u00ec hay truy c\u1eadp ngay v\u00e0o website: Vietlearn.org\/ c\u1ee7a ch\u00fang t\u00f4i \u0111\u1ec3 t\u00ecm hi\u1ec3u nh\u00e9!<\/p>\n\n\n\n Xem ngay:<\/p>\n\n\n\n Chia \u0111\u01a1n th\u1ee9c cho \u0111\u01a1n th\u1ee9c \u2013 H\u1ecdc t\u1ed1t to\u00e1n 8 c\u00f9ng Vietlearn<\/p>\n\n\n\n Nh\u1eefng h\u1eb3ng \u0111\u1eb3ng th\u1ee9c \u0111\u00e1ng<\/p>\n\n\n\n\n\n\n","protected":false},"excerpt":{"rendered":" Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn Nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c l\u00e0 d\u1ea1ng to\u00e1n \u0111\u1ea7u ti\u00ean m\u00e0 c\u00e1c b\u1ea1n h\u1ecdc sinh ph\u1ea3i chinh ph\u1ee5c n\u1ebfu nh\u01b0 mu\u1ed1n h\u1ecdc t\u1ed1t m\u00f4n to\u00e1n. Tuy l\u00e0 d\u1ea1ng to\u00e1n \u0111\u01a1n gi\u1ea3n nh\u01b0ng n\u1ebfu kh\u00f4ng t\u1eadp trung t\u00ednh to\u00e1n r\u1ea5t d\u1ec5 x\u1ea3y ra l\u1ed7i sai. V\u00ec v\u1eady h\u00f4m nay, […]<\/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-973","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\/973","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=973"}],"version-history":[{"count":0,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/973\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/media?parent=973"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/categories?post=973"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/tags?post=973"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}} = x3+ 2x<\/code><\/pre>\n\n\n\n= x.x2 \u2013 x.y \u2013 (x2.x + x2.y) + y.x2 \u2013 y.x\n\n= x3 \u2013 xy \u2013 x3 \u2013 x2y + x2y \u2013 xy\n\n= (x3 \u2013 x3) + (x2y \u2013 x2y) \u2013 xy \u2013 xy\n\n= \u20132xy<\/code><\/pre>\n\n\n\n= x.x2 \u2013 x.y \u2013 (x2.x + x2.y) + y.x2 \u2013 y.x\n\n= x3 \u2013 xy \u2013 x3 \u2013 x2y + x2y \u2013 xy\n\n= (x3 \u2013 x3) + (x2y \u2013 x2y) \u2013 xy \u2013 xy\n\n= \u20132xy<\/code><\/pre>\n\n\n\n