\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)

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

a/ Đơn giản là dùng phép thế:

\(x+2y+x+y+z=0\Rightarrow x+2y=0\Rightarrow x=-2y\)

\(x+y+z=0\Rightarrow z=-\left(x+y\right)=-\left(-2y+y\right)=y\)

Thế vào pt cuối:

\(\left(1-2y\right)^2+\left(y+2\right)^2+\left(y+3\right)^2=26\)

Vậy là xong

b/ Sử dụng hệ số bất định:

\(\left\{{}\begin{matrix}a\left(\frac{x}{3}+\frac{y}{12}-\frac{z}{4}\right)=a\\b\left(\frac{x}{10}+\frac{y}{5}+\frac{z}{3}\right)=b\end{matrix}\right.\)

\(\Rightarrow\left(\frac{a}{3}+\frac{b}{10}\right)x+\left(\frac{a}{12}+\frac{b}{5}\right)y+\left(\frac{-a}{4}+\frac{b}{3}\right)z=a+b\) (1)

Ta cần a;b sao cho \(\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}=-\frac{a}{4}+\frac{b}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}\\\frac{a}{3}+\frac{b}{10}=-\frac{a}{4}+\frac{b}{3}\end{matrix}\right.\) \(\Rightarrow\frac{a}{2}=\frac{b}{5}\)

Chọn \(\left\{{}\begin{matrix}a=2\\b=5\end{matrix}\right.\) thay vào (1):

\(\frac{7}{6}\left(x+y+z\right)=7\Rightarrow x+y+z=6\)

b, Ta có : \(\left\{{}\begin{matrix}x+y+z=3\\y+z+t=4\\z+t+x=5\\t+x+y=6\end{matrix}\right.\)

=> \(x+y+z+y+z+t+z+t+x+t+x+y=18\)

=> \(3\left(x+y+z+t\right)=18\)

=> \(x+y+z+t=6\)

=> \(x+y+z+t=x+y+t\)

=> \(z=0\)

=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\x+y+t=6\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\y+5=6\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+1=3\\t+1=4\\x+t=5\\y=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2\\t=3\\x+t=5\\y=1\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=1\\z=0\\t=3\end{matrix}\right.\)

a, Ta có : \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\9yz=20\left(y+z\right)\\8zx=15\left(z+x\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}7xy-12x-12y=0\\9yz-20y-20z=0\\8zx-15z-15x=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{12y}{7y-12}\\y=\frac{20z}{9z-20}\\x=\frac{15z}{8z-15}\end{matrix}\right.\)

=> \(12y\left(8z-15\right)=15z\left(7y-12\right)\)

=> \(96yz-180y=105yz-180z\)

=> \(105yz-96yz=-180y+180z\)

=> \(9yz=-180y+180z\)

=> \(180z-180y=20y+20z\)

=> \(180z-20z=180y+20y=160z=200y\)

=> \(y=\frac{4}{5}z\)

=> \(\frac{20z}{9z-20}=\frac{4z}{5}\)

=> \(4z\left(9z-20\right)=100z\)

=> \(36z^2-180z=0\)

=> \(\left[{}\begin{matrix}z=5\\z=0\end{matrix}\right.\)

TH1 : z = 0 .

=> \(x=y=z=0\)

TH₂ : z = 5 .

=> \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\45y=20\left(y+5\right)\\40x=15\left(5+x\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\)