Giải hệ \(\left\{{}\begin{matrix}x^2y+2x^2+3y-15=0\\x^4+y^2-2x^2-4y-5=0\end{matrix}\right.\)
Giải hệ \(\left\{{}\begin{matrix}xy+3y^2-x+4y=7\\2xy+y^2-2x-2y+1=0\end{matrix}\right.\)
giải hệ phương trình
1)\(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\) 2)\(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\) 3)\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}-x+3y=16\\2x+y=3\end{matrix}\right.\) 5)\(\left\{{}\begin{matrix}\dfrac{-3}{x-y}+\dfrac{5}{2x+y}=-2\\\dfrac{4}{x-y}-\dfrac{10}{2x+y}=2\end{matrix}\right.\) 6)\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
giải hpt sau
\(\left\{{}\begin{matrix}3x^2+6xy-x+3y=0\\4x-9y=6\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)
giải hệ phương trình
a
\(\left\{{}\begin{matrix}x+y=1\\x-y=-5\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}2x+2y=5\\x-2y=1\end{matrix}\right.\)
c.
\(\left\{{}\begin{matrix}2x+3y=5\\3x-2y=1\end{matrix}\right.\)
giải hệ phương trình
a) \(\left\{{}\begin{matrix}x+2y=2\\-2x+y=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}2y-x=2\\2x-y=-1\end{matrix}\right.\)
giúp tui giải bài này với tui c.ơn trước
Giải hệ bằng phương pháp phân tích đa thức thành nhân tử
a) \(\left\{{}\begin{matrix}xy+x-2=0\\2x^3-x^2y+x^2+y^2-2xy-y=0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2-2xy+x+y=0\\x^4-4x^2y+3x^2+y^2=0\end{matrix}\right.\)
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)\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3-3xy^2-2y^3=0\\3xy+2x-4y=6\end{matrix}\right.\)giải hệ pt sau