Question

Giải hệ phương trình sau:

\(\left\{{}\begin{matrix}\frac{8}{x-3}+\frac{1}{2\left|y\right|-3}=5\\\frac{4}{x-3}+\frac{1}{2\left|y\right|-3}=3\end{matrix}\right.\)

Answer 1

$\begin{cases}\dfrac{8}{x-3}+\dfrac{1}{2|y|-3}=5\\\dfrac{4}{x-3}+\dfrac{1}{2|y|-3}=3\end{cases}\,\,\Bigg(x\ne3;\,y\ne\pm\dfrac{3}{2}\Bigg)\\\text{Đặt }\dfrac{1}{x-3}=a;\,\dfrac{1}{2|y|-3}=b\\\to \begin{cases}8a+b=5\\4a+b=3\end{cases}\to \begin{cases}4a=2\\4a+b=3\end{cases}\\\to \begin{cases}a=\dfrac{1}{2}\\4.\dfrac{1}{2}+b=3\end{cases}\to \begin{cases}a=\dfrac{1}{2}\\b=1\end{cases}\\\to \begin{cases}\dfrac{1}{x-3}=\dfrac{1}{2}\\\dfrac{1}{2|y|-3}=1\end{cases}\to \begin{cases}x-3=2\\2|y|-3=1\end{cases}\\\to \begin{cases}x=5\\2|y|=4\end{cases} \to \begin{cases}x=5\\|y|=2\end{cases}\\\to \begin{cases}x=5\\y=\pm2\end{cases}\text{ (thoả mãn)}\\\to\text{Hệ có nghiệm }(x;\,y)=(5;\,2);\,(x;\,y)=(5;\,-2)$