Question

giải hệ pt sau:

\(\left\{{}\begin{matrix}\dfrac{14}{x-y+2}-\dfrac{10}{x+y-1}=9\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)

Answer 1

Lời giải:
Đặt $\frac{1}{x-y+2}=a;\frac{1}{x+y-1}=b$ thì HPT trở thành cơ bản:
\(\left\{\begin{matrix} 14a-10b=9\\ 3a+2b=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 14a-10b=9\\ 15a+10b=20\end{matrix}\right.\)

$\Rightarrow (14a-10b)+(15a+10b)=9+20$

$\Leftrightarrow 29a=29\Leftrightarrow a=1$.

$b=\frac{4-3a}{2}=\frac{1}{2}$

Vậy: \(\left\{\begin{matrix} \frac{1}{x-y+2}=1\\ \frac{1}{x+y-1}=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-y+2=1\\ x+y-1=2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x-y=-1\\ x+y=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\end{matrix}\right.\)