Giải các hệ PT sau:
a) \(\left\{{}\begin{matrix}2x^2-3xy=y^2-3x-1\\2y^2-3xy=x^2-3y-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^3-2y=4\\y^3-2x=4\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\sqrt{x+1}-\sqrt{7-y}=4\\\sqrt{y+1}-\sqrt{7-x}=4\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}2x^2=y+\frac{1}{y}\\2y^2=x+\frac{1}{x}\end{matrix}\right.\)
Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left|x-2\right|+2\left|y-1\right|=9\\x+\left|y-1\right|=-1\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\\\sqrt[3]{4x+y+1}+\sqrt{3x+2y}=4\end{matrix}\right.\)
Giải hệ phương trình:
1, \(\left\{{}\begin{matrix}x^2+1+y^2+xy=y\\x+y-2=\frac{y}{1+x^2}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+8y^3-4xy^2=1\\2x^4+8y^4-2x-y=0\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x^2+y^2=\frac{1}{5}\\4x^2+3x-\frac{57}{25}=-y\left(3x+1\right)\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\sqrt{12-y}+\sqrt{y\left(12-x\right)}=12\\x^3-8x-1=2\sqrt{y-2}\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
câu 3: giải hệ phương trình
a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{5x}{6}-y=\dfrac{-5}{6}\\\dfrac{2x}{2x+y}+3y=\dfrac{-2}{3}\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x\sqrt{3}+3y=1\\2x-y\sqrt{3}=\sqrt{3}\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.=17\)
giúp mk vs ạ mk cần gấp
1) Giải hệ phương trình:
(1) \(\left\{{}\begin{matrix}5x\sqrt{3}+y=2\sqrt{2}\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\)
Giải hệ phương trình sau: \(\left\{{}\begin{matrix}\sqrt{3+2x^2y-x^4y^2}+x^2\left(1-2x^2\right)=y^4\\1+\sqrt{1+\left(x-y\right)^2}+x^2\left(x^4-2x^2-2xy^2+1\right)=0\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}2x^2-3xy=y^2-3x-1\\2y^2-3xy=x^2-3y-1\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}\dfrac{1}{3x}+\dfrac{2x}{3y}=\dfrac{x+\sqrt{y}}{2x^2+y}\\2\left(2x+\sqrt{y}\right)=\sqrt{2x+6}-y\end{matrix}\right.\)