Toán Lớp 9: Giải hệ phương trình : 2 $\sqrt[]{x+1}$ – $\frac{3}{y-1}$ =1
$\sqrt[]{x+1}$ + $\frac{1}{y-1}$ =3
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Toán Lớp 9: Giải hệ phương trình : 2 $\sqrt[]{x+1}$ – $\frac{3}{y-1}$ =1
$\sqrt[]{x+1}$ + $\frac{1}{y-1}$ =3
Comments ( 1 )
Dkxd:\left\{ \begin{array}{l}
x + 1 \ge 0\\
y – 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge – 1\\
y \ne 1
\end{array} \right.\\
Dat:\sqrt {x + 1} = a;\dfrac{1}{{y – 1}} = b\\
\Leftrightarrow \left\{ \begin{array}{l}
2a – 3b = 1\\
a + b = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2a – 3b = 1\\
2a + 2b = 6
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
5b = 5\\
a = 3 – b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 1\\
a = 3 – 1 = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{{y – 1}} = 1\\
\sqrt {x + 1} = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y – 1 = 1\\
x + 1 = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 2\left( {tm} \right)\\
x = 3\left( {tm} \right)
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {3;2} \right)
\end{array}$