\(ĐK:x\ne-2;y\ne4\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{y-1}-\dfrac{3}{x+2}=\dfrac{9}{4}\left(1\right)\\\dfrac{5}{y-1}+\dfrac{3}{x+2}=\dfrac{29}{12}\left(2\right)\end{matrix}\right.\\ \left(1\right)+\left(2\right)=\dfrac{14}{y-1}=\dfrac{14}{3}\\ \Leftrightarrow y-1=3\\ \Leftrightarrow y=4\left(tm\right)\\ \text{Thay vào }\left(2\right)\Leftrightarrow\dfrac{5}{3}+\dfrac{3}{x+2}=\dfrac{29}{12}\\ \Leftrightarrow\dfrac{3}{x+2}=\dfrac{3}{4}\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
Vậy \(\left(x;y\right)=\left(2;4\right)\)
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