Giải hpt sau:
a)\(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}+7=0\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\sqrt{4x^2-8x+4}+5\sqrt{y^2+4y+4}=13\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
Ghpt:\(\left\{{}\begin{matrix}\left(4x^2+1\right).x+\left(y-3\right)\sqrt{5-2y}=0\\4x^2+y^2+2\sqrt{3-4x}=7\end{matrix}\right.\)
GHPT: \(\left\{{}\begin{matrix}\left(4x^2+1\right).x+\left(y-3\right)\sqrt{5-2y}=0\\4x^2+y^2+2\sqrt{3-4x}=7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4y^3-12y^2+13y-5=\left(4x+9\right)\sqrt{x+2}\\2\left(x^2-5\left(y-1^2\right)\right)=3\left(y-1\right)\sqrt{x^2-4x-8}\end{matrix}\right.\)
Giải hệ bằng phương pháp phân tích nhân tử
a) \(\left\{{}\begin{matrix}x^2+2y=xy+4\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2-2xy+x+y=0\\x^4-4x^2y+3x^2+y^2=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)x+2y=1\\4x-\left(\sqrt{2}+1\right)y=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)x+2y=1\\4x-\left(\sqrt{2}+1\right)y=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\sqrt{4x^2-8x+4}+5\sqrt{y^2+4y+4}=13\end{matrix}\right.\)
Giải hệ \(\left\{{}\begin{matrix}x^2+xy\left(2y-1\right)=2y^3-2y^2-x\\6\sqrt{x-1}+y+7=4x\left(y-1\right)\end{matrix}\right.\)