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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
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 \(\Leftrightarrow \left\{\begin{matrix} x(x+y+z)=2\\ y(y+z+x)=3\\ z(z+x+y)=4\end{matrix}\right.(*)\).
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 \(k=\frac{-1}{3}\Rightarrow (x,y,z)=(2k,3k,4k)=(\frac{-2}{3}; -1; \frac{-4}{3})\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y+1=2\\yz+y+z+1=5\\zx+z+x+1=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=2\\\left(y+1\right)\left(z+1\right)=5\\\left(z+1\right)\left(x+1\right)=10\end{matrix}\right.\) (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 \(y>0\)
\(\left\{{}\begin{matrix}x+y+z=6\left(1\right)\\xy+yz-zx=7\\x^2+y^2+z^2=14\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\left(x+y+z\right)^2=36\\xy+yz-xz=7\\x^2+y^2+z^2=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+xz\right)=36\\xy+yz-xz=7\\x^2+y^2+z^2=14\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}14+2\left(xy+yz+xz\right)=36\\xy+yz-xz=7\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}xy+yz+xz=11\\xy+yz-xz=7\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}xy+yz=\frac{11+7}{2}=9\\xz=\frac{11-7}{2}=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y\left(x+z\right)=9\\x=\frac{2}{z}\end{matrix}\right.\)
=>\(y\left(\frac{2}{z}+z\right)=9\)
<=> \(y=\frac{9}{\frac{2}{z}+z}=\frac{9}{\frac{2+z^2}{z}}=\frac{9z}{2+z^2}\)
Thay \(x=\frac{2}{z},y=\frac{9z}{2+z^2}\) vào (1) có:
\(\frac{2}{z}+\frac{9z}{2+z^2}+z=6\)
<=> \(\frac{2\left(2+z^2\right)+9z^2+z^2\left(2+z^2\right)}{z\left(2+z^2\right)}=6\)
<=>\(4+2z^2+9z^2+2z^2+z^4=6z\left(2+z^2\right)\)
<=> \(z^4+13z^2+4-12z-6z^3=0\)
<=> \(z^4-3z^3+2z^2-3z^3+9z^2-6z+2z^2-6z+4=0\)
<=>\(z^2\left(z^2-3z+2\right)-3z\left(z^2-3z+2\right)+2\left(z^2-3z+2\right)=0\)
<=> \(\left(z^2-3z+2\right)^2=0\)
<=> \(z^2-3z+2=0\)<=> \(z\left(z-2\right)-\left(z-2\right)=0\)
<=> \(\left(z-1\right)\left(z-2\right)=0\)
=>\(\left[{}\begin{matrix}z=1\\z=2\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\frac{2}{z}=2,y=\frac{9z}{2+z^2}=3\\x=1,y=3\end{matrix}\right.\)
Vậy (x,y,z) \(\in\left\{\left(2,3,1\right),\left(1,3,2\right)\right\}\)
Cộng vế:
\(\Rightarrow x^2+y^2+2xy+x+y=20\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)-20=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=4\\x+y=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=4-x\\y=-5-x\end{matrix}\right.\)
Thế vào pt đầu...
