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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