Ta có : \(\left\{{}\begin{matrix}5-4y=20\\\frac{x}{4}-\frac{y}{5}=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-\frac{15}{4}\\\frac{x}{4}-\frac{-\frac{15}{4}}{5}=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-\frac{15}{4}\\\frac{x}{4}=\frac{7}{4}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-\frac{15}{4}\\x=7\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(\left(x,y\right)=\left(7,-\frac{15}{4}\right)\)
Giải hpt sau:
a) \(\left\{{}\begin{matrix}\sqrt{5}x+\left(1-\sqrt{3}\right)y=1\\\left(1-\sqrt{3}\right)x+\sqrt{5}y=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{3x}{x+1}-\frac{2y}{y+4}=4\\\frac{2x}{x+1}-\frac{5y}{y+4}=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{2}{2x-y}+\frac{3}{x-2y}=\frac{1}{2}\\\frac{2}{2x-y}-\frac{1}{x-2y}=\frac{1}{18}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{2x+3}{x+1}+\frac{4}{\sqrt{y}-2}=4\\\frac{5}{x+1}-\frac{2}{\sqrt{y}-2}=4\end{matrix}\right.\)
\(\frac{7}{x-y+2}-\frac{5}{x+y-1}=\frac{9}{2}\)
\(\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\)
giải hpt:
\(\left\{{}\begin{matrix}\frac{5}{\sqrt{x-1}}-3\left|y+1\right|=-4\\\frac{2}{\sqrt{x-1}}+\left|y+1\right|=5\end{matrix}\right.\)
Giải hộ mình câu này với ạ
\(\left\{{}\begin{matrix}\frac{2}{x-1}\frac{1}{y+1}=7\\\frac{5}{x-1}-\frac{2}{y-1}=4\end{matrix}\right.\)
giả hpt
\(\left\{{}\begin{matrix}3\sqrt{x-1}-\frac{1}{3}\sqrt{y+1}=5\\5\sqrt{x-1}-2\sqrt{y+1}=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}2\left(x+y\right)=5\left(x-y\right)\\\frac{20}{x+y}+\frac{20}{x-y}=7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{3}{y-2}=4\\\frac{4}{x}+\frac{1}{y-2}=1\end{matrix}\right.\)
