Bài 1 trang 139 Tài liệu dạy – học Toán 9 tập 2

Đề bài

Giải các hệ phương trình:

a) \(\left\{ \begin{array}{l}x + y = 2\\2x – 3y = 9\end{array} \right.\)

b) \(\left\{ \begin{array}{l}\dfrac{x}{y} = \dfrac{2}{3}\\x + y – 10 = 0\end{array} \right.\)

c) \(\left\{ \begin{array}{l}3(x – y) – y = 11\\x – 2(x + 5y) =  – 15\end{array} \right.\)

d) \(\left\{ \begin{array}{l}\dfrac{2}{{2x – y}} + \dfrac{3}{{x – 2y}} = \dfrac{1}{2}\\\dfrac{2}{{2x – y}} – \dfrac{1}{{x – 2y}} = \dfrac{1}{{18}}\end{array} \right.\)

Lời giải chi tiết

\(a)\,\,\left\{ \begin{array}{l}x + y = 2\\2x – 3y = 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}2x + 2y = 4\\2x – 3y = 9\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}5y =  – 5\\x + y = 2\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y =  – 1\\x = 2 – \left( { – 1} \right) = 3\end{array} \right.\)

Vậy nghiệm \(\left( {x;y} \right)\) của hệ phương trình là \(\left( {3; – 1} \right)\).

b)

\(\left\{ \begin{array}{l}\dfrac{x}{y} = \dfrac{2}{3}\\x + y – 10 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3x – 2y = 0\\x + y – 10 = 0\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}3x – 2y = 0\\2x + 2y = 20\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}5x = 20\\y = 10 – x\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = 4\\y = 10 – 4 = 6\end{array} \right.\)

Vậy nghiệm \(\left( {x;y} \right)\) của hệ phương trình là \(\left( {4;6} \right)\).

\(\begin{array}{l}c)\,\,\left\{ \begin{array}{l}3(x – y) – y = 11\\x – 2(x + 5y) =  – 15\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}3x – 3y – y = 11\\x – 2x – 10y =  – 15\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}3x – 4y = 11\\ – x – 10y =  – 15\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3x – 4y = 11\\ – 3x – 30y =  – 45\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} – 34y =  – 34\\3x – 4y = 11\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y = 1\\3x – 4.1 = 11\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y = 1\\3x = 15\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}x = 5\\y = 1\end{array} \right.\end{array}\)

Vậy nghiệm \(\left( {x;y} \right)\) của hệ phương trình là \(\left( {5;1} \right)\).

\(d)\,\,\left\{ \begin{array}{l}\dfrac{2}{{2x – y}} + \dfrac{3}{{x – 2y}} = \dfrac{1}{2}\\\dfrac{2}{{2x – y}} – \dfrac{1}{{x – 2y}} = \dfrac{1}{{18}}\end{array} \right.\)

Đặt \(\left\{ \begin{array}{l}\dfrac{1}{{2x – y}} = u\\\dfrac{1}{{x – 2y}} = v\end{array} \right.\,\,\left( {u,v \ne 0} \right)\). Khi đó hệ phương trình trở thành:

\(\begin{array}{l}\left\{ \begin{array}{l}2u + 3v = \dfrac{1}{2}\\2u – v = \dfrac{1}{{18}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}4v = \dfrac{4}{9}\\2u – v = \dfrac{1}{{18}}\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}v = \dfrac{1}{9}\\2u – \dfrac{1}{9} = \dfrac{1}{{18}}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}v = \dfrac{1}{9}\\u = \dfrac{1}{{12}}\end{array} \right.\,\,\left( {tm} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{1}{{2x – y}} = \dfrac{1}{{12}}\\\dfrac{1}{{x – 2y}} = \dfrac{1}{9}\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}2x – y = 12\\x – 2y = 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}2x – y = 12\\2x – 4y = 18\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3y =  – 6\\x – 2y = 9\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}y =  – 2\\x – 2\left( { – 2} \right) = 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y =  – 2\\x = 5\end{array} \right.\end{array}\).

Vậy nghiệm \(\left( {x;y} \right)\) của hệ phương trình là \(\left( {5; – 2} \right)\).

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