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\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)
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
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 :))
\(hpt\Leftrightarrow\left\{{}\begin{matrix}x+y+xy+1=4\\y+z+yz+1=2\\x+z+xz+1=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=4\\\left(y+1\right)\left(z+1\right)=2\\\left(x+1\right)\left(z+1\right)=2\end{matrix}\right.\)
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....
\(\left\{{}\begin{matrix}xy+y+x+1=10\\yz+y+z+1=5\\zx+x+z+1=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=10\\\left(y+1\right)\left(z+1\right)=5\\\left(z+1\right)\left(x+1\right)=2\end{matrix}\right.\)
\(\Rightarrow\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=100\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=10\)
\(\Rightarrow\left\{{}\begin{matrix}z+1=1\\x+1=2\\y+1=5\end{matrix}\right.\)
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} x(x+y+z)=2\\ y(y+z+x)=3\\ z(z+x+y)=4\end{matrix}\right.(*)\).
Dễ thấy $x+y+z\neq 0$. Khi đó ta có:
\(\frac{x}{y}=\frac{x(x+y+z)}{y(y+z+x)}=\frac{2}{3}(1)\)
\(\frac{y}{z}=\frac{y(y+z+x)}{z(z+x+y)}=\frac{3}{4}(2)\)
Từ \((1);(2)\Rightarrow \frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) .
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\Rightarrow x=2k; y=3k; z=4k\)
Thay vào PT thứ nhất của $(*)$ suy ra:
\(2k(2k+3k+4k)=2\)
\(\Leftrightarrow 18k^2=2\Rightarrow k=\pm \frac{1}{3}\)
Nếu \(k=\frac{1}{3}\Rightarrow (x,y,z)=(2k,3k,4k)=(\frac{2}{3}; 1; \frac{4}{3})\)
Nếu \(k=\frac{-1}{3}\Rightarrow (x,y,z)=(2k,3k,4k)=(\frac{-2}{3}; -1; \frac{-4}{3})\)
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
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y+1=2\\yz+y+z+1=5\\zx+z+x+1=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=2\\\left(y+1\right)\left(z+1\right)=5\\\left(z+1\right)\left(x+1\right)=10\end{matrix}\right.\) (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 \(y>0\)