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

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

Aki Tsuki hattori heiji Akai Haruma

b: =>x^2-y^2-4y-2x-3=0 và x^2+2x+y=0

=>x^2-2x+1-y^2-4y-4=0 và x^2+2x+y=0

=>x=1 và y=-2 và x^2+2x+y=0

=>Hệ vô nghiệm

a: \(\Leftrightarrow\left\{{}\begin{matrix}z=2x-5\\y=3-2x+z=3-2x+2x-5=-2\\3x-2\cdot\left(-2\right)+2x-5=14\end{matrix}\right.\)

=>y=-2; 3x+4+2x-5=14; z=2x-5

=>y=-2; x=3; z=2*3-5=1

b) Áp dụng bđt Svac-xơ:

\(\dfrac{1}{x}+\dfrac{9}{y}+\dfrac{16}{z}\ge\dfrac{\left(1+3+4\right)^2}{x+y+z}\ge\dfrac{64}{4}=16>9\)

=> hpt vô nghiệm

c) Ở đây x,y,z là các số thực dương

Áp dụng cosi: \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)=3xyz\)

Dấu = xảy ra khi \(x=y=z=\dfrac{3}{3}=1\)

Đặt S=x+y;P=xy giải ra :V

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 .

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

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

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