Giải hệ phương trình 5x+ $\frac{2}{|y-1|}$ =7 2x- $\frac{1}{|1-y|}$ = 1

Đáp án: $\,\left( {x;y} \right) = \left( {1;2} \right)/\left( {1;0} \right)$

Giải thích các bước giải:

$\begin{array}{l}
Dkxd:y \ne 1\\
\left\{ \begin{array}{l}
5x + \dfrac{2}{{\left| {y – 1} \right|}} = 7\\
2x – \dfrac{1}{{\left| {y – 1} \right|}} = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
5x + \dfrac{2}{{\left| {y – 1} \right|}} = 7\\
4x – \dfrac{2}{{\left| {y – 1} \right|}} = 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
9x = 9\\
2x – \dfrac{1}{{\left| {y – 1} \right|}} = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 1\\
\dfrac{1}{{\left| {y – 1} \right|}} = 2x – 1 = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 1\\
\left| {y – 1} \right| = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 1\\
\left[ \begin{array}{l}
y – 1 = 1\\
y – 1 =  – 1
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 1\\
\left[ \begin{array}{l}
y = 2\\
y = 0
\end{array} \right.
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {1;2} \right)/\left( {1;0} \right)
\end{array}$