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

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

ta có : x+xy+y=1

<=> x(y+1) + (y+1)=2

<=> (x+1)(y+1)=2

tương tự(y+1)(z+1)=5

(x+1)(z+1)=10

ta đc hệ pt............

đặt x+1=a,y+1=b,z+1=c

ta có : ab=2 (1) , bc=5 (2) , ac=10

=> abc=2c , abc=5a , abc= 10b

=> 5a=10b=2c

+ 5a=10b

=> a=2b . (1)=> 2b^2=1=> b=1 hoặc b=-1

=> a=2 hoăc a=-2 . (2)=> c=5 hoăc c=-5

like nha :))

PT (1) \(\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

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

Nhận thấy VT\(\ge\)0 với mọi x,y,z

Dấu = xảy ra <=> x=y=z

Thay x=y=z vào pt (2) ta được:

\(3x^{2021}=3^{2022}\) \(\Leftrightarrow x^{2021}=3^{2021}\) \(\Leftrightarrow x=3\)

\(\Rightarrow x=y=z=3\)

Vậy (x;y;z)=(3;3;3)

Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}=\frac{3}{8}\\ \frac{1}{y}+\frac{1}{z}=\frac{3}{4}\\ \frac{1}{z}+\frac{1}{x}=\frac{5}{6}\end{matrix}\right.\Rightarrow 2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{8}+\frac{3}{4}+\frac{5}{6}\)

\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{47}{48}\)

\(\Rightarrow \left\{\begin{matrix} \frac{1}{z}=\frac{47}{48}-\frac{3}{8}\\ \frac{1}{x}=\frac{47}{48}-\frac{3}{4}\\ \frac{1}{y}=\frac{47}{48}-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{48}{29}\\ y=\frac{48}{11}\\ z=\frac{48}{7}\end{matrix}\right.\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}x+y+xy+1=4\\y+z+yz+1=2\\x+z+xz+1=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=4\\\left(y+1\right)\left(z+1\right)=2\\\left(x+1\right)\left(z+1\right)=2\end{matrix}\right.\)

Lấy \(\dfrac{pt\left(2\right)}{pt\left(3\right)}\Leftrightarrow\dfrac{y+1}{x+1}=1\)\(\Leftrightarrow y+1=x+1\)\(\Leftrightarrow x=y\)

Thay vào \(pt(1)\)\(\Leftrightarrow x^2+2x=3\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=y=1\\x=y=-3\end{matrix}\right.\)

Thay vào \(pt\left(3\right)\)\(\Leftrightarrow\left[{}\begin{matrix}z+1+z=1\\z-3-3z=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}z=0\\z=-2\end{matrix}\right.\)

Vậy....

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=\dfrac{9}{9}=1\)

Dau bang xay ra khi x=y=z=3 ( vi x+y+z=9)

\(\left\{{}\begin{matrix}xy+y+x+1=10\\yz+y+z+1=5\\zx+x+z+1=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=10\\\left(y+1\right)\left(z+1\right)=5\\\left(z+1\right)\left(x+1\right)=2\end{matrix}\right.\)

\(\Rightarrow\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=100\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=10\)

\(\Rightarrow\left\{{}\begin{matrix}z+1=1\\x+1=2\\y+1=5\end{matrix}\right.\)