đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)
\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)
\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)
\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)
\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)
Giải giùm mình bài này
$\left[\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{y-x}\right].\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}$
\(\)
\(\hept{\begin{cases}\left(x-2\right)\left(y+3\right)=5+xy\\x\left(y-3\right)=xy\end{cases}}\)
\(\hept{\begin{cases}\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=4\\\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{y}}=6\end{cases}}\)
GIải giúp mình 2 hệ này với :<
CMR với mọi x, y: \(\left(x+y\right)^2-xy+1\ge\left(x+y\right)\sqrt{3}\)
CÁC BẠN ƠI GIẢI GIÚP MÌNH BÀI NÀY VỚI
Giúp mình giải bài toán này với.
\(\hept{\begin{cases}\left(x+2\right)^2+4\left(y-1\right)^2=4xy+13\\\sqrt{\frac{x^2-xy-2y^2}{x-y}}+\sqrt{x+y}=\frac{2}{\sqrt{x^2-y^2}}\end{cases}}\)
câu 1: giải hệ phương trình
\(\left(x+y\right)^2+\left(y+z\right)^4+....+\left(x+z\right)^{100}=-\left(y+z+x\right)\)
\(\left(xy\right)^2+2\left(yz\right)^4+....+100\left(zx\right)^{100}=-[\left(x+y+z\right)+2\left(yz+zx+xy\right)+......+99\left(x+y+z\right)]\)\(\left(\frac{1}{x}+\frac{1}{y}\right)^2+\left(\frac{1}{y^2}+\frac{1}{z^2}\right)^2+...+\left(\frac{1}{x^{99}}+\frac{1}{z^{99}}\right)^2=-\frac{1}{\left(xy\right)^2+2\left(yz\right)^2+.....+99\left(zx\right)^2}\)
tìm x,y,z
giải hệ phương trình:
1) \(\hept{\begin{cases}2\left(x+y\right)+3\left(x+y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{cases}}\)
2)\(\hept{\begin{cases}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12_{ }\end{cases}}\)
3) \(\hept{\begin{cases}\frac{2y-5x}{3}+5=\frac{y+27}{4}-2x\\\frac{x+1}{3}+y=\frac{6y-5x}{7}\end{cases}}\)
4)\(\hept{\begin{cases}\frac{1}{2}\left(x+2\right)\left(y+3\right)-\frac{1}{2}xy=50\\\frac{1}{2}xy-\frac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{cases}}\)
5)\(\hept{\begin{cases}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{cases}}\)
\(\sqrt{x^2+xy+y^2}=\sqrt{\left(x+y\right)^2-xy}\ge\sqrt{\left(x+y\right)^2-\frac{1}{4}\left(x+y\right)^2}=\frac{x+y}{2}.\sqrt{3}\)
cmtt=>\(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)=3\)
\(\hept{\begin{cases}\left(x+y\right)^2-\left(x+y\right)\sqrt{3}-xy=-1\\x^2+y^2+x+2y=\sqrt{3}+\frac{2}{3}\end{cases}}\)
Giúp mình giải he pt này với