Giải hệ phương trình : 2 $\sqrt[]{x+1}$ - $\frac{3}{y-1}$ =1 $\sqrt[]{x+1}$ + $\frac{1}{y-1}$ =3

Đáp án: $\left( {x;y} \right) = \left( {3;2} \right)$

Giải thích các bước giải:

$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x + 1 \ge 0\\
y - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
y \ne 1
\end{array} \right.\\
Dat:\sqrt {x + 1}  = a;\dfrac{1}{{y - 1}} = b\\
 \Leftrightarrow \left\{ \begin{array}{l}
2a - 3b = 1\\
a + b = 3
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2a - 3b = 1\\
2a + 2b = 6
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
5b = 5\\
a = 3 - b
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
b = 1\\
a = 3 - 1 = 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{{y - 1}} = 1\\
\sqrt {x + 1}  = 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y - 1 = 1\\
x + 1 = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = 2\left( {tm} \right)\\
x = 3\left( {tm} \right)
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {3;2} \right)
\end{array}$