Điều kiện là các số dương
\(VT=\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(VT\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\frac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(VT\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\ge\frac{8}{9}\left(x+y+z\right).3\sqrt[3]{x^2y^2z^2}=VP\)
Dấu "=" xảy ra khi \(x=y=z\)
giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
Cho x+y+z=0 và x,y,z khác 0. Rút gọn:
a) A= \(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) B= \(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
Cho x, y, z là các số thực thuộc (0;1) thỏa mãn điều kiện \(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\)
Tìm giá trị lớn nhất của biểu thức \(P=\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+3xy-\left(x^2+y^2\right)\)
Cho x3+y3+z3=3xyz. Rút gọn:
P=\(\frac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Giải hpt: \(\left\{{}\begin{matrix}\left(x+y\right)\left(y+z\right)=187\\\left(y+z\right)\left(z+x\right)=154\\\left(z+x\right)\left(x+y\right)=238\end{matrix}\right.\)
(x, y, z > 0)
1,\(\left\{{}\begin{matrix}x^2-2y^2-xy=0\\\sqrt{2x}+\sqrt{y+1}=2\end{matrix}\right.\)
2,\(\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+y^2\right)=x\left(y+1\right)\\\sqrt{x}+\sqrt{y+1}=2\end{matrix}\right.\)
3,\(\left\{{}\begin{matrix}2y^3-\left(x+4\right)y^2+8y+x^2-4x=0\\\sqrt{\frac{1-x}{2}}+\sqrt{x+2y+3}=\sqrt{5}\end{matrix}\right.\)
Giải hệ phương trình sau :
\(\begin{cases}xy\left(x+1\right)=x^3+y^2+x-y\\3y\left(2+\sqrt{9x^2+3}\right)+\left(4y+2\right)\left(\sqrt{1+x+x^2}+1\right)=0\end{cases}\) \(\left(x,y\in Z\right)\)
hệ phương trình
1 ,\(\left\{{}\begin{matrix}\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{3}{2}\\3x-2y=5\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{x^2-y-6}{x}=x-2\\x+3y=8\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{2}{3}\\x+y=10\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\frac{y^2+2x-8}{y}=y-3\\x+y=10\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}\frac{x+1}{y-1}=5\\3\left(2x-2\right)-4\left(3x+4\right)=5\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}2x+y=4\\\left|x-2y\right|=3\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{2x}{x+1}+\frac{y}{y+1}=3\\\frac{x}{x+1}-\frac{3y}{y+1}=-1\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}y-\left|x\right|=1\\2x-y=1\end{matrix}\right.\)
10 , \(\left\{{}\begin{matrix}\sqrt{x+3y}=\sqrt{3x-1}\\5x-y=9\end{matrix}\right.\)
