Đặt \(\dfrac{1}{x+1}=a;\dfrac{1}{y-5}=b\)
\(\left\{{}\begin{matrix}a-b=1\\5a+3b=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-3b=3\\5a+3b=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{4}\\b=-\dfrac{3}{4}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=\dfrac{11}{3}\end{matrix}\right.\)
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