giúp mik giải bài hệ pt vs ạ!
1,\(\left\{{}\begin{matrix}x^2+y^2+\dfrac{2xy}{x+y}=1\\\sqrt{x+y}=x^2-y\end{matrix}\right.\)
2,\(\left\{{}\begin{matrix}2x^3+xy^2+x=y^3+4x^2y+2y\\\sqrt{4x^2+x+6}-5\sqrt{1+2y}=1-4y\end{matrix}\right.\)
3,\(\left\{{}\begin{matrix}2x^2+\sqrt{2}x=\left(x+y\right)y+\sqrt{x+y}\\\sqrt{x-1}+xy=\sqrt{y^2+21}\end{matrix}\right.\)
4,\(\left\{{}\begin{matrix}\sqrt{9y^2+\left(2y+3\right)\left(y-x\right)}+4\sqrt{xy}=7x\\\left(2y-1\right)\sqrt{1+x}+\left(2y+1\right)\sqrt{1-x}=2y\end{matrix}\right.\)
Giải hệ pt
\(\left\{{}\begin{matrix}3\sqrt{2x+y}+\sqrt{x-2y+1}=5\\2\sqrt{x-2y+1}-5x=10y+9\end{matrix}\right.\)
a) Giải pt: \(x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)
b)Giải hệ pt \(\left\{{}\begin{matrix}xy-y^2+2y-x-1=\sqrt{y-1}-\sqrt{x}\\3\sqrt{6-y}+3\sqrt{2x+3y-7}=2x+7\end{matrix}\right.\)
Giải hệ pt : \(\left\{{}\begin{matrix}\sqrt{x^2-\left(x+y\right)}=\frac{y}{\sqrt[3]{x-y}}\\2\left(x^2+y^2\right)-3\sqrt{2x-1}=11\end{matrix}\right.\)
giải hệ phương trình:
\(\left\{{}\begin{matrix}\sqrt{7x+y}+\sqrt{2x+y}=5\\\sqrt{x+4y}+x-y=2\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}\sqrt{4-x}+\sqrt{y+8}=y^2+7x-1\\\sqrt{2\left(x-y\right)^2+6y-2x+4}-\sqrt{x}=\sqrt{y+1}\end{matrix}\right.\)
giải hệ pt:
\(\left\{{}\begin{matrix}\sqrt{2x+y-1}-\sqrt{x+2y-2}+x-y+1=0\\4x^2-y^2+x+4=\sqrt{2x+y}+\sqrt{x+4y}\end{matrix}\right.\)
giải hệ pt:
\(\left\{{}\begin{matrix}\sqrt{x^2-x-y}=\frac{y}{\sqrt[3]{x-y}}\\2\left(x^2+y^2\right)-2\sqrt{2x-1}=13\end{matrix}\right.\)
Giải hệ phương trình:
\(\left\{{}\begin{matrix}\sqrt{4-x}+\sqrt{y+8}=y^2+7x-1\\\sqrt{2\left(x-y\right)^2+6y-2x+4}-\sqrt{x}=\sqrt{y+1}\end{matrix}\right.\)