Vì \(|x-2|\)nằm ở mẫu \(\Rightarrow\)x - 2 > 0 \(\Rightarrow\) x > 2 \(\Rightarrow\) \(|x-2|\)= x-2
\(\left\{{}\begin{matrix}\frac{2}{x-2}+\frac{1}{y}=2\\\frac{6}{x-2}-\frac{2}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{6}{x-2}+\frac{3}{y}=6\\\frac{6}{x-2}-\frac{2}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{y}=6\\\frac{6}{x-2}-\frac{2}{y}=1\end{matrix}\right.\)
giải hệ: a, \(\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\sqrt[]{x-1}+\sqrt[]{y-1}=2\\\frac{1}{x}+\frac{1}{y}=1\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x\sqrt[]{x}+y\sqrt[]{y}=35\\x\sqrt[]{y}+y\sqrt[]{x}=30\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x^2+xy+y^2=3\\x+xy+y=-1\end{matrix}\right.\)
e,\(\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\)
giải hệ phương trình
\(\left\{{}\begin{matrix}\frac{1}{x}-\frac{3}{y-2}=2\\\frac{2}{x}-\frac{1}{2-y}=11\end{matrix}\right.\)
giải hệ phương trình
\(\left\{{}\begin{matrix}x+\left|y-12\right|=0\\-x+2y=27\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{5x-9}{x-2}=11\\\frac{3x-4}{x-2}=-11\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}x+3y+5=xy\\\frac{1}{x-1}+\frac{1}{y-2}=2\end{matrix}\right.\)
giải hệ phương trình \(\left\{{}\begin{matrix}\frac{2x}{x-1}+\sqrt{y+1}=7\\\frac{3x}{x-1}-2\sqrt{y+1}=0\end{matrix}\right.\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}\frac{4}{x+y-1}-\frac{5}{2x-y+3}=\frac{-5}{2}\\\frac{3}{x+y-1}+\frac{1}{2x-y+3}=\frac{-7}{5}\end{matrix}\right.\)
Giải hệ phương trình :
\(\left\{{}\begin{matrix}3x^3-y^3=\frac{1}{x+y}\\x^2+y^2=1\end{matrix}\right.\)
giải hệ phương trình
\(\left\{{}\begin{matrix}8\left(x^3-1\right)+6xy^2=y\left(12x^2+y^2\right)\\\left(x^2+y-4x\right)\left(x^2-y^2-2x-5\right)=14\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{x+1}{3}-\frac{y+2}{4}=\frac{2\left(x-y\right)}{5}\\\frac{x-3}{4}-\frac{y-3}{3}=2y-x\end{matrix}\right.\)