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

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

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

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

Nhân vế với vế: \(\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=100\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=10\) (2)

Chia vế cho vế của (2) cho từng pt của (1):

\(\Rightarrow\left\{{}\begin{matrix}z+1=5\\x+1=2\\y+1=1\end{matrix}\right.\) \(\Rightarrow\left(x;y;z\right)=\left(1;0;4\right)\) (loại)

Hệ vô nghiệm do \(y>0\)

3(x² + y² + x²) = 3[(x + y + z)² - 2(xy + yz + zx)] = 3(9 + 2) = 33

Pt thứ 3 tương đương với pt:

x³ + y³ + z³ + 6 = 33

<=> x³ + y³ + z³ = 27 = (x + y + z)³

<=> (x + y + z)³ - x³ - y³ - z³ = 0

<=> 3(x + y)(y + z)(z + x) = 0

Đến đây khá dễ rồi, tự làm tiếp nhé

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 :))

a: \(\Leftrightarrow\left\{{}\begin{matrix}2x+2y+4z=8\\2x-y+3z=6\\2x-6y+8z=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y+z=2\\8y-4z=1\\x+y+2z=4\end{matrix}\right.\)

=>y=9/20; z=13/20; x=4-y-2z=9/4

b: \(\Leftrightarrow\left\{{}\begin{matrix}z=23-x-y\\z=31-y-t\\z=27-t-x\\x+y+t=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x-y+23=-y-t+31\\-y-t-31=-x-t+27\\x+y+t=33\\z=23-x-y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x+t=8\\x-y=58\\x+y+t=33\\z=23-x-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=x+8\\y=x-58\\x-58+x+8+x=33\\z=23-x-y\end{matrix}\right.\)

=>x=83/3; t=107/3; y=-91/3; z=23-83/3+91/3=77/3