Giải hệ phương trình sau:
a. \(\left\{{}\begin{matrix}\dfrac{x+2}{y}=\dfrac{x+1}{y-2}\\\dfrac{5x+1}{5x-2}=\dfrac{y-2}{y+2}\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}2x+\left|y\right|=4\\4x-3y=1\end{matrix}\right.\)
Giai he phuong trinh bang phuong phap cong va phuong phap the
<=> \(\left\{{}\begin{matrix}4x+3x=-6\\\dfrac{x+3y}{3}-\dfrac{y-2}{5}=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+2y=10\\\dfrac{1}{2}x-y=1\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{1}{y}=2\\\dfrac{6}{x}-\dfrac{2}{y}=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2}{\left|x-2\right|}+\dfrac{1}{y}=2\\\dfrac{6}{\left|x-2\right|}-\dfrac{2}{y}=1\end{matrix}\right.\)
Giải hpt
a)\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y+12}-\dfrac{x}{y}=2\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y}{xy}+\dfrac{xy}{x+y}=\dfrac{5}{2}\\\dfrac{x-y}{xy}+\dfrac{xy}{x-y}=\dfrac{10}{3}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{2y}{x-1}-\dfrac{5x}{y-1}=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}+\dfrac{4}{\left|y\right|-2}=\dfrac{11}{6}\\\dfrac{5}{\left|x+3\right|}+\dfrac{2}{\left|y\right|-2}=\dfrac{11}{6}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)=\dfrac{1}{2}xy+56\\\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=\dfrac{1}{2}xy-32\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{\left|x+2\right|}+\dfrac{1}{\sqrt{y-2}}=4\\\dfrac{2}{\left|x+2\right|}-\dfrac{1}{\sqrt{y-2}}=1\end{matrix}\right.\)
Giải các hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2y-5x}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2\left(x-1\right)+\dfrac{3}{y+2}=5\\\left(x-1\right)-\dfrac{1}{y+2}=\dfrac{5}{3}\end{matrix}\right.\)