xin slot tối làm =)
xin slot tối làm =)
\(\left\{{}\begin{matrix}\frac{2}{x-2}+3\sqrt{y+1}=4\\\frac{4}{x-2}-\sqrt{y+1}=1\end{matrix}\right.\) tìm đk rồi giải hpt hộ mình nha
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) \(\left\{{}\begin{matrix}\frac{2}{2x-y}+\frac{3}{x-2y}=\frac{1}{2}\\\frac{2}{2x-y}-\frac{1}{x-2y}=\frac{1}{18}\end{matrix}\right.\)
giải hpt:
\(\left\{{}\begin{matrix}\frac{5}{\sqrt{x-1}}-3\left|y+1\right|=-4\\\frac{2}{\sqrt{x-1}}+\left|y+1\right|=5\end{matrix}\right.\)
giả hpt
\(\left\{{}\begin{matrix}3\sqrt{x-1}-\frac{1}{3}\sqrt{y+1}=5\\5\sqrt{x-1}-2\sqrt{y+1}=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{2x+3}{x+1}+\frac{4}{\sqrt{y}-2}=4\\\frac{5}{x+1}-\frac{2}{\sqrt{y}-2}=4\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}\frac{1}{1+\left(x-y\right)^2}=z+4\\\sqrt{z+3}+2x=8\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{2x+1}-\frac{y}{3y+2}=1\\3\sqrt{2x+1}-\frac{5y}{3y+2}=1\end{matrix}\right.\)
Giải các hệ phương trình sau:
1, \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\dfrac{1}{x-5}+\dfrac{6}{\sqrt{y}-2}=2\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x+7}}-\dfrac{4}{\sqrt{y-6}}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x+7}}+\dfrac{3}{\sqrt{y-6}}=\dfrac{13}{6}\end{matrix}\right.\)
Tìm điều kiện giúp mình nhé!