Giải hệ pt:
\(\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)\left(x+y-2\right)=6\\x^2+y^2-2x-2y=3\end{matrix}\right.\)
giải hệ pt:
(1)\(\left\{{}\begin{matrix}2\text{x}+2y+2\text{x}y=10\\x^2+y^2=5\end{matrix}\right.\)
(2)\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=3\\\sqrt{xy}=2\end{matrix}\right.\)
(3)\(\left\{{}\begin{matrix}x-y=1\\x.y=6\end{matrix}\right.\)
(4)\(\left\{{}\begin{matrix}|x|+y=3\\2|x|-y=3\end{matrix}\right.\)
giải hệ pt:
(1) \(\left\{{}\begin{matrix}x^2-3xy+2y^2=0\\3x+y=6\end{matrix}\right.\)
(2)\(\left\{{}\begin{matrix}\dfrac{x-1}{2x+1}-\dfrac{y-2}{y+2}=1\\\dfrac{3x-3}{2x+1}+\dfrac{2y-4}{y+2}=3\end{matrix}\right.\)
(3)\(\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}=4\\x+y-3\sqrt{x+1}=-5\end{matrix}\right.\)
Giải hệ pt
a) \(\left\{{}\begin{matrix}x^3+6x^2y=7\\2y^3+3xy^2=5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}6x-xy-2=0\\2\sqrt{\left(x+2\right)\left(3x-y\right)}=y+6\end{matrix}\right.\)
giải các hệ pt sau:
a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\left|x-1\right|+y=5\\4x+3y=23\end{matrix}\right.\)
Giải hệ pt:
\(\left\{{}\begin{matrix}x^3+y^3+3xy=1\\\sqrt{\left(4-x\right)\left(13-y\right)}=\dfrac{2x+2y+25}{2x+y+2}\end{matrix}\right.\)
Giải hệ pt:
\(\left\{{}\begin{matrix}x^2+2y^2+3xy+3=0\\\dfrac{x-y+18}{\left(x+y\right)^2}=9\sqrt{x-y}\end{matrix}\right.\)
giải hệ pt\(\left\{{}\begin{matrix}x^3+y^3-3xy=-1\\x^2y+y^2x+x^2+y^2=4\end{matrix}\right.\)
giải giúp mình mấy hệ pt sau nhé
1.\(\left\{{}\begin{matrix}2x^3+y\left(x+10=4x^2\right)\\5x^4-4x^6=y^2\end{matrix}\right.\)
2.\(\left\{{}\begin{matrix}x-\sqrt{y+1}=\dfrac{5}{2}\\y+\left(x-3\right)\sqrt{x+1}=-\dfrac{3}{4}\end{matrix}\right.\)
3.\(\left\{{}\begin{matrix}x^2+y^2=2\\\left(x+y\right)\left(4-x^2y^2-2xy\right)=2y^2,2y^3,2y^5\end{matrix}\right.\)trong 3 cái đó có 1 cái nha
4.\(\left\{{}\begin{matrix}x^3-8x=y^3+2y\\x^2-3y^2=6\end{matrix}\right.\)
mọi người giúp nhanh nha
thanks nhiều