\(ĐK:x\ne-2;y\ne\dfrac{3}{2}\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x+2}+\dfrac{2}{2y-3}=4\left(1\right)\\\dfrac{6}{x+2}-\dfrac{2}{2y-3}=1\left(2\right)\end{matrix}\right.\\ \text{Lấy }\left(1\right)+\left(2\right)=\dfrac{10}{x+2}=5\\ \Leftrightarrow x+2=2\\ \Leftrightarrow x=0\left(tm\right)\\ \text{Thay vào PT}\left(1\right)\Leftrightarrow1+\dfrac{1}{2y-3}=2\\ \Leftrightarrow\dfrac{1}{2y-3}=1\\ \Leftrightarrow2y-3=1\Leftrightarrow y=2\left(tm\right)\)
Vậy \(\left(x;y\right)=\left(0;2\right)\)
Đúng 1
Bình luận (0)
