giải hệ 1 \(\left\{{}\begin{matrix}6xy=5\left(x+y\right)\\3yz=2\left(y+z\right)\\7zx=10\left(z+x\right)\end{matrix}\right.\)
2.\(\left\{{}\begin{matrix}xy-x-y=5\\yz-y-z=11\\zx-z-x=7\end{matrix}\right.\)
3.\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\\y^2+xy-yz+z^2=0\\x^2-xy-xz-z^2=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)
Lấy (2) cộng (3) ta được
\(x^2+y^2-yz-zx=2\) (4)
Lấy (1) - (4) ta được
\(2x\left(x+z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-z\end{matrix}\right.\)
Xét 2 TH rồi thay vào tìm được y và z
1. \(\left\{{}\begin{matrix}6xy=5\left(x+y\right)\\3yz=2\left(y+z\right)\\7zx=10\left(z+x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{6}{5}\\\dfrac{y+z}{yz}=\dfrac{3}{2}\\\dfrac{z+x}{zx}=\dfrac{7}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{5}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{3}{2}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{7}{10}\end{matrix}\right.\)
Đến đây thì dễ rồi nhé
2. \(\left\{{}\begin{matrix}\left(xy-x\right)-\left(y-1\right)=6\\\left(yz-y\right)-\left(z-1\right)=12\\\left(zx-z\right)-\left(x-1\right)=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)=6\\\left(y-1\right)\left(z-1\right)=12\\\left(z-1\right)\left(x-1\right)=8\end{matrix}\right.\)
Đến đây dễ rồi