giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y+z}=\dfrac{1}{2}\\\dfrac{1}{y}+\dfrac{1}{z+x}=\dfrac{1}{3}\\\dfrac{1}{z}+\dfrac{1}{x+y}=\dfrac{1}{4}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6}\\x^2-y^2=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{5}{\sqrt{x-7}}+\dfrac{3}{\sqrt{y+6}}=\dfrac{13}{6}\\\dfrac{7}{\sqrt{x-7}}-\dfrac{2}{\sqrt{y+6}}=\dfrac{5}{3}\end{matrix}\right.\)
giải hệ phương trình nghiệm nguyên sau:\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(y+\dfrac{1}{y}\right)\\y=\dfrac{1}{2}\left(z+\dfrac{1}{z}\right)\\z=\dfrac{1}{2}\left(x+\dfrac{1}{x}\right)\end{matrix}\right.\)
Giai hệ phương trình:
\(\left\{{}\begin{matrix}x+y+z=\dfrac{1}{2}\\\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{1}{xyz}=4\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}>0\end{matrix}\right.\)
Giải hệ phương trình:
\(\left\{{}\begin{matrix}x+y+z=9\\\sqrt{x}+\sqrt{y}+\sqrt{z}=5\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{3}{2}\end{matrix}\right.\)
Giải hệ phương trình:
\(\left\{{}\begin{matrix}x^2\left(y-z\right)=-\dfrac{5}{3}\left(1\right)\\y^2\left(z-x\right)=3\left(2\right)\\z^2\left(x-y\right)=\dfrac{1}{3}\left(3\right)\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}1+\left(x+z\right)^2=\dfrac{1}{y^2}-\sqrt{y-1}\\y=\dfrac{x^2+y^2}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x^2}{1+x^2}=y\\\dfrac{2y^2}{1+y^2}=z\\\dfrac{2z^2}{1+z^2}=x\end{matrix}\right.\)
giả các hệ phương trình sau :
a) \(\left\{{}\begin{matrix}\dfrac{-3}{x-y+1}+\dfrac{1}{x +y-2}=12\\\dfrac{2}{x-y+1}-\dfrac{3}{x+y-2}=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+2\left(y^2+2y\right)=10\\3x^2-\left(y^2+2y\right)=9\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x-1}}-\dfrac{5}{\sqrt{y+2}}=\dfrac{9}{2}\\\dfrac{3}{\sqrt{x-1}}+\dfrac{2}{\sqrt{y+2}}=4\end{matrix}\right.\)
Giải hệ phương trình sau (bằng cách cộng đại số):
3/ \(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\)