giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 ,...
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giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
Trừ theo vế hai pt đầu của hệ:
(x-y)(x+y-z)=0\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x+y=z\end{matrix}\right.\)
Xét x=y. Khi đó ta có hệ mới:\(\left\{{}\begin{matrix}y^2+yz=4\\z^2+y^2=10\end{matrix}\right.\)
=>5y2+5yz=2z2+2y2<=>3y2+5yz-2z2=0<=>\(\left[{}\begin{matrix}y=\frac{1}{3}z\\y=-2z\end{matrix}\right.\)
y=-2z=>(-2z)2-2z.z=4<=>2z2=4<=>\(\left[{}\begin{matrix}z=\sqrt{2}\rightarrow x=y=-2\sqrt{2}\\z=-\sqrt{2}\rightarrow x=y=2\sqrt{2}\end{matrix}\right.\)
\(y=\frac{1}{3}z\Rightarrow\left(\frac{1}{3}z\right)^2+\frac{1}{3}z.z=4\Leftrightarrow z^2=9\Leftrightarrow\left[{}\begin{matrix}z=3\rightarrow x=y=1\\z=-3\rightarrow x=y=-1\end{matrix}\right.\)
Xét x+y=z. Cộng theo vế hai pt đầu:
x2+y2+(x+y)2=8
=>4[(x+y)2+xy]=5[(x+y)2+x2+y2]<=>3x2-xy+3y2=0(pt vô nghiệm)