Giải hệ phương trình \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=5\\\frac{2}{xy}-\frac{1}{z^2}-20=5\end{matrix}\right.\)
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giải hệ phương trình
1 , left{{}begin{matrix}left(x+yright)left(x-1right)left(x-yright)left(x+1right)+2xyleft(y-xright)left(y-1right)left(y+xright)left(y-2right)-2xyend{matrix}right.
2, left{{}begin{matrix}2left(frac{1}{x}+frac{1}{2y}right)+3left(frac{1}{x}-frac{1}{2y}right)^29left(frac{1}{x}+frac{1}{2y}right)-6left(frac{1}{x}-frac{1}{2y}right)^2-3end{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...
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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 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
giải hệ phương trình \(\left\{{}\begin{matrix}y=x^2\\\frac{1}{x}=\frac{1}{y}+\frac{1}{z}\\z=xy\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=x^2\\\frac{1}{x}=\frac{1}{y}+\frac{1}{z}\\z=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\\frac{1}{x}=\frac{1}{y}+\frac{1}{xy}\\z=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\\frac{1}{x}=\frac{x}{xy}+\frac{1}{xy}=\frac{x+1}{xy}\\z=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\xy=x^2+x\\z=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\x^3-x^2-x=0\\z=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\x\left(x^2-x-1\right)=0\\z=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\left(loại\right)\\\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\\y=x^2\\z=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\frac{\sqrt{5}+1}{2}\left(TM\right)\\x=\frac{1-\sqrt{5}}{2}\left(TM\right)\end{matrix}\right.\\y=x^2\\z=xy\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\frac{1+\sqrt{5}}{2}\left(\right)TM\\y=\frac{3+\sqrt{5}}{2}\left(TM\right)\\z=2+\sqrt{5}\left(TM\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=\frac{1-\sqrt{5}}{2}\\y=\frac{3-\sqrt{5}}{2}\left(TM\right)\\z=2-\sqrt{5}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)
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Giaỉ hệ phương trình \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\\\frac{2}{xy}-\frac{1}{z^2}=4\end{matrix}\right.\)
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\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=2-\frac{1}{z}\\\frac{2}{xy}=4+\frac{1}{z^2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4+\frac{1}{z^2}-\frac{4}{z}\\\frac{2}{xy}=4+\frac{1}{z^2}\end{matrix}\right.\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}\) (1)
Từ pt đầu suy ra:
\(\frac{1}{x}+\frac{1}{y}-2=-\frac{1}{z}\Rightarrow\frac{4}{x}+\frac{4}{y}-8=-\frac{4}{z}\) (2)
Thế (2) vào (1)
\(\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{x}+\frac{4}{y}-8\)
\(\Leftrightarrow\frac{1}{x^2}-\frac{4}{x}+4+\frac{1}{y^2}-\frac{4}{y}+4=0\)
\(\Leftrightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\)
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Giải hệ phương trình: \(\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=\frac{9}{2}\\xy+\frac{1}{xy}=\frac{5}{2}\end{matrix}\right.\)
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Giải hệ phương trình
\(\left\{{}\begin{matrix}x\left(yz+1\right)=\frac{7}{3}z\\y\left(xz+1\right)=8x\\z\left(xy+1\right)=\frac{9}{2}y\end{matrix}\right.\)
Giải các hệ phương trình sau
a)left{{}begin{matrix}frac{1}{x}+frac{1}{y+1}12x+3yxy+5end{matrix}right.
b)left{{}begin{matrix}left(x-yright)^2+3left(x-yright)42x+3y12end{matrix}right.
c)left{{}begin{matrix}frac{x}{y}+frac{y}{x}frac{13}{6}x+y5end{matrix}right.
d)left{{}begin{matrix}x+y+xy7x+y^2+xy13end{matrix}right.
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Giải các hệ phương trình sau
a)\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\2x+3y=xy+5\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(x-y\right)^2+3\left(x-y\right)=4\\2x+3y=12\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\frac{x}{y}+\frac{y}{x}=\frac{13}{6}\\x+y=5\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}x+y+xy=7\\x+y^2+xy=13\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}x+y^2+z^3=14\\\left(\frac{1}{2x}+\frac{1}{3y}+\frac{1}{6z}\right)\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{6}\right)=1\end{matrix}\right.\)
giải hệ phương trình
\(\left\{{}\begin{matrix}y=x^2\\\frac{1}{x}=\frac{1}{y}+\frac{6}{z}\\z=xy\end{matrix}\right.\) với x,y,z\(\ne\)0
Giải hệ phương trình \(\left\{{}\begin{matrix}x+y+z=a\\x^2+y^2+z^2=b^2\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\end{matrix}\right.\)