{"id":34185,"date":"2026-10-26T11:46:34","date_gmt":"2026-10-26T04:46:34","guid":{"rendered":"https:\/\/etest.edu.vn\/?p=34185"},"modified":"2026-10-26T11:46:34","modified_gmt":"2026-10-26T04:46:34","slug":"cach-giai-rational-equations-trong-sat-math-va-tranh-nghiem-ngoai-lai","status":"publish","type":"post","link":"https:\/\/etest.edu.vn\/en\/cach-giai-rational-equations-trong-sat-math-va-tranh-nghiem-ngoai-lai\/","title":{"rendered":"C\u00e1ch gi\u1ea3i rational equations trong SAT Math v\u00e0 tr\u00e1nh nghi\u1ec7m ngo\u1ea1i lai"},"content":{"rendered":"

\u0110i\u1ec1u ng\u01b0\u1eddi h\u1ecdc th\u01b0\u1eddng g\u1eb7p<\/b><\/h1>\n

Ph\u01b0\u01a1ng tr\u00ecnh h\u1eefu t\u1ec9 (Rational Equations) c\u00f3 th\u1ec3 khi\u1ebfn h\u1ecdc sinh e ng\u1ea1i v\u00ec ch\u1ee9a c\u00e1c ph\u00e2n th\u1ee9c v\u00e0 y\u00eau c\u1ea7u nhi\u1ec1u b\u01b0\u1edbc bi\u1ebfn \u0111\u1ed5i. Tuy nhi\u00ean, th\u1eed th\u00e1ch l\u1edbn nh\u1ea5t kh\u00f4ng ch\u1ec9 n\u1eb1m \u1edf vi\u1ec7c quy \u0111\u1ed3ng hay nh\u00e2n ch\u00e9o, m\u00e0 c\u00f2n \u1edf kh\u1ea3 n\u0103ng x\u00e1c \u0111\u1ecbnh \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh v\u00e0 ki\u1ec3m tra nghi\u1ec7m sau khi gi\u1ea3i. Khi n\u1eafm v\u1eefng quy tr\u00ecnh gi\u1ea3i v\u00e0 h\u00ecnh th\u00e0nh th\u00f3i quen \u0111\u1ed1i chi\u1ebfu k\u1ebft qu\u1ea3 v\u1edbi \u0111i\u1ec1u ki\u1ec7n ban \u0111\u1ea7u, h\u1ecdc sinh c\u00f3 th\u1ec3 x\u1eed l\u00fd d\u1ea1ng b\u00e0i n\u00e0y m\u1ed9t c\u00e1ch ch\u00ednh x\u00e1c v\u00e0 t\u1ef1 tin h\u01a1n trong Digital SAT Math.<\/span><\/p>\n

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C\u00e1ch gi\u1ea3i rational equations trong SAT Math v\u00e0 tr\u00e1nh nghi\u1ec7m ngo\u1ea1i lai<\/figcaption><\/figure>\n

Rational equation l\u00e0 g\u00ec?<\/b><\/h1>\n

Rational expression (bi\u1ec3u th\u1ee9c h\u1eefu t\u1ec9) l\u00e0 th\u01b0\u01a1ng c\u1ee7a hai \u0111a th\u1ee9c, v\u00ed d\u1ee5 (x + 1)\/(x \u2212 2). Rational equation l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 \u00edt nh\u1ea5t m\u1ed9t bi\u1ec3u th\u1ee9c h\u1eefu t\u1ec9 ch\u1ee9a \u1ea9n \u1edf m\u1eabu, v\u00ed d\u1ee5:<\/span><\/p>\n

V\u00ec kh\u00f4ng \u0111\u01b0\u1ee3c chia cho 0, m\u1ecdi gi\u00e1 tr\u1ecb l\u00e0m m\u1eabu s\u1ed1 b\u1eb1ng 0 \u0111\u1ec1u b\u1ecb lo\u1ea1i kh\u1ecfi t\u1eadp x\u00e1c \u0111\u1ecbnh. \u0110\u00e2y l\u00e0 \u0111i\u1ec3m kh\u00e1c bi\u1ec7t c\u1ed1t l\u00f5i so v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh \u0111a th\u1ee9c th\u00f4ng th\u01b0\u1eddng.<\/span><\/p>\n

\u0110\u1eb7c t\u1ea3 c\u1ee7a College Board x\u1ebfp ph\u01b0\u01a1ng tr\u00ecnh h\u1eefu t\u1ec9 v\u00e0o nh\u00f3m “nonlinear equations in one variable” thu\u1ed9c m\u1ea3ng Advanced Math, c\u00f9ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai, ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a c\u0103n v\u00e0 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i. Trong \u0111\u1ec1 th\u1eadt, d\u1ea1ng n\u00e0y c\u00f3 th\u1ec3 l\u00e0 c\u00e2u tr\u1eafc nghi\u1ec7m ho\u1eb7c c\u00e2u t\u1ef1 \u0111i\u1ec1n \u0111\u00e1p \u00e1n.<\/span><\/p>\n

Quy tr\u00ecnh 4 b\u01b0\u1edbc solving rational equations<\/b><\/h1>\n

B\u01b0\u1edbc 1. T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh. Ph\u00e2n t\u00edch t\u1eebng m\u1eabu th\u00e0nh nh\u00e2n t\u1eed v\u00e0 ghi ra c\u00e1c gi\u00e1 tr\u1ecb l\u00e0m m\u1eabu b\u1eb1ng 0. V\u00ed d\u1ee5 x\u00b2 \u2212 1 = (x \u2212 1)(x + 1) cho \u0111i\u1ec1u ki\u1ec7n x \u2260 1 v\u00e0 x \u2260 \u22121.<\/span><\/p>\n

B\u01b0\u1edbc 2. T\u00ecm m\u1eabu s\u1ed1 chung nh\u1ecf nh\u1ea5t. M\u1eabu s\u1ed1 chung nh\u1ecf nh\u1ea5t (least common denominator) ch\u00ednh l\u00e0 b\u1ed9i chung nh\u1ecf nh\u1ea5t c\u1ee7a c\u00e1c m\u1eabu: l\u1ea5y m\u1ed7i nh\u00e2n t\u1eed v\u1edbi s\u1ed1 m\u0169 l\u1edbn nh\u1ea5t (c\u00e1ch t\u00ecm \u0111\u01b0\u1ee3c tr\u00ecnh b\u00e0y trong b\u00e0i “Least Common Multiple l\u00e0 g\u00ec?”).<\/span><\/p>\n

B\u01b0\u1edbc 3. Nh\u00e2n hai v\u1ebf v\u1edbi m\u1eabu s\u1ed1 chung \u0111\u1ec3 kh\u1eed m\u1eabu. Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh ph\u01b0\u01a1ng tr\u00ecnh \u0111a th\u1ee9c, th\u01b0\u1eddng l\u00e0 b\u1eadc nh\u1ea5t ho\u1eb7c b\u1eadc hai.<\/span><\/p>\n

B\u01b0\u1edbc 4. Gi\u1ea3i v\u00e0 \u0111\u1ed1i chi\u1ebfu \u0111i\u1ec1u ki\u1ec7n. Nghi\u1ec7m n\u00e0o vi ph\u1ea1m \u0111i\u1ec1u ki\u1ec7n \u1edf b\u01b0\u1edbc 1 l\u00e0 extraneous solution (nghi\u1ec7m ngo\u1ea1i lai) v\u00e0 ph\u1ea3i lo\u1ea1i.<\/span><\/p>\n

C\u00f3 nghi\u1ec7m ngo\u1ea1i lai<\/b><\/h1>\n

\u0110\u01b0\u1ee3c x = \u22122 ho\u1eb7c x = 1. Nh\u01b0ng x = 1 l\u00e0m m\u1eabu b\u1eb1ng 0 n\u00ean l\u00e0 nghi\u1ec7m ngo\u1ea1i lai. Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m duy nh\u1ea5t x = \u22122.<\/span><\/p>\n

\u0110\u1ec1 SAT th\u01b0\u1eddng \u0111\u1eb7t c\u1ea3 hai gi\u00e1 tr\u1ecb \u22122 v\u00e0 1 (ho\u1eb7c t\u1ed5ng c\u1ee7a ch\u00fang) v\u00e0o c\u00e1c ph\u01b0\u01a1ng \u00e1n. N\u1ebfu qu\u00ean lo\u1ea1i nghi\u1ec7m ngo\u1ea1i lai, em s\u1ebd ch\u1ecdn \u0111\u00fang ph\u01b0\u01a1ng \u00e1n nhi\u1ec5u. B\u01b0\u1edbc ph\u00e2n t\u00edch x\u00b2 + x \u2212 2 th\u00e0nh nh\u00e2n t\u1eed \u0111\u01b0\u1ee3c luy\u1ec7n k\u1ef9 trong b\u00e0i “Factoring Quadratic Trinomials”.<\/span><\/p>\n

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Quy tr\u00ecnh 4 b\u01b0\u1edbc solving rational equations<\/figcaption><\/figure>\n

Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<\/b><\/h1>\n

D\u1ea1ng “ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 bao nhi\u00eau nghi\u1ec7m” r\u1ea5t hay xu\u1ea5t hi\u1ec7n \u1edf c\u00e2u tr\u1eafc nghi\u1ec7m v\u1edbi c\u00e1c ph\u01b0\u01a1ng \u00e1n: kh\u00f4ng c\u00f3 nghi\u1ec7m, \u0111\u00fang m\u1ed9t nghi\u1ec7m, hai nghi\u1ec7m, v\u00f4 s\u1ed1 nghi\u1ec7m. Ng\u01b0\u1eddi kh\u00f4ng \u0111\u1ed1i chi\u1ebfu \u0111i\u1ec1u ki\u1ec7n s\u1ebd tr\u1ea3 l\u1eddi “\u0111\u00fang m\u1ed9t nghi\u1ec7m”.<\/span><\/p>\n

T\u00ecm tham s\u1ed1 \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<\/b><\/h1>\n

\u0110\u00e2y l\u00e0 phi\u00ean b\u1ea3n kh\u00f3 h\u01a1n c\u1ee7a d\u1ea1ng 4, th\u01b0\u1eddng n\u1eb1m \u1edf n\u1eeda sau module 2. \u0110\u1ec1 cho ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a h\u1eb1ng s\u1ed1 a v\u00e0 h\u1ecfi a b\u1eb1ng bao nhi\u00eau th\u00ec ph\u01b0\u01a1ng tr\u00ecnh kh\u00f4ng c\u00f3 nghi\u1ec7m.<\/span><\/p>\n

Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh sau khi kh\u1eed m\u1eabu r\u01a1i \u0111\u00fang v\u00e0o gi\u00e1 tr\u1ecb l\u00e0m m\u1eabu b\u1eb1ng 0 (nh\u01b0 v\u00ed d\u1ee5 5).<\/span><\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh sau khi kh\u1eed m\u1eabu t\u1ef1 n\u00f3 v\u00f4 nghi\u1ec7m, t\u1ee9c c\u00f3 d\u1ea1ng 0 \u00b7 x = c v\u1edbi c \u2260 0.<\/span><\/p>\n

Ki\u1ec3m tra b\u1eb1ng Desmos<\/b><\/h1>\n

V\u1edbi v\u00ed d\u1ee5 3, hai \u0111\u1ed3 th\u1ecb c\u1eaft nhau t\u1ea1i x = \u22122. T\u1ea1i x = 1, c\u1ea3 hai h\u00e0m \u0111\u1ec1u kh\u00f4ng x\u00e1c \u0111\u1ecbnh n\u00ean kh\u00f4ng c\u00f3 giao \u0111i\u1ec3m, v\u00e0 nghi\u1ec7m ngo\u1ea1i lai t\u1ef1 bi\u1ebfn m\u1ea5t. Ng\u01b0\u1ee3c l\u1ea1i, n\u1ebfu em ch\u1ec9 v\u1ebd ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai sau khi kh\u1eed m\u1eabu (y = x\u00b2 + x \u2212 2), Desmos s\u1ebd hi\u1ec7n c\u1ea3 hai nghi\u1ec7m v\u00e0 em l\u1ea1i r\u01a1i v\u00e0o b\u1eaby. H\u00e3y lu\u00f4n v\u1ebd ph\u01b0\u01a1ng tr\u00ecnh g\u1ed1c, \u0111\u1eebng v\u1ebd ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 bi\u1ebfn \u0111\u1ed5i. C\u00e1c thao t\u00e1c c\u01a1 b\u1ea3n v\u1edbi Desmos c\u00f3 trong b\u00e0i “Desmos l\u00e0 g\u00ec?”.<\/span><\/p>\n

L\u1ed7i th\u01b0\u1eddng g\u1eb7p khi gi\u1ea3i rational equations<\/b><\/h1>\n

Ch\u1ec9 nh\u00e2n m\u1ed9t v\u1ebf v\u1edbi m\u1eabu chung: ph\u1ea3i nh\u00e2n m\u1ecdi h\u1ea1ng t\u1eed \u1edf c\u1ea3 hai v\u1ebf, k\u1ec3 c\u1ea3 h\u1ea1ng t\u1eed kh\u00f4ng c\u00f3 m\u1eabu nh\u01b0 s\u1ed1 2 trong v\u00ed d\u1ee5 4.<\/span><\/p>\n

Nh\u00e2n ch\u00e9o khi ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nhi\u1ec1u h\u01a1n hai ph\u00e2n th\u1ee9c: nh\u00e2n ch\u00e9o ch\u1ec9 d\u00f9ng cho d\u1ea1ng A\/B = C\/D. V\u1edbi t\u1ed5ng nhi\u1ec1u ph\u00e2n th\u1ee9c, quay v\u1ec1 quy tr\u00ecnh 4 b\u01b0\u1edbc.<\/span><\/p>\n

R\u00fat g\u1ecdn nh\u00e2n t\u1eed c\u00f3 ch\u1ee9a \u1ea9n m\u00e0 kh\u00f4ng ghi \u0111i\u1ec1u ki\u1ec7n: r\u00fat g\u1ecdn (x \u2212 2)\/(x \u2212 2) th\u00e0nh 1 nh\u01b0ng qu\u00ean r\u1eb1ng x \u2260 2.<\/span><\/p>\n

Sai d\u1ea5u khi ph\u00e2n ph\u1ed1i: \u22121 \u00b7 (x \u2212 2) = \u2212x + 2. Xem th\u00eam c\u00e1c b\u1eaby d\u1ea5u trong b\u00e0i “Negative numbers l\u00e0 g\u00ec?”.<\/span><\/p>\n

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L\u1ed7i th\u01b0\u1eddng g\u1eb7p khi gi\u1ea3i rational equations<\/figcaption><\/figure>\n

C\u00e1ch t\u1ef1 ki\u1ec3m tra khi luy\u1ec7n<\/b><\/h1>\n