\(x-\sqrt{xy}+y=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}\sqrt{y}+\dfrac{1}{4}y+\dfrac{3}{4}y\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y>0\)
\(\sqrt{xy}>0\)
Do đó: D>0
=>\(D>\sqrt{D}\)
Cho \(H=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)( \(x>y\ge0\)). So sánh H với \(\sqrt{H}\)
Cho \(H=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\) \(\left(x>y\ge0\right)\). So sánh H với\(\sqrt{H}\)
Cho x,y >0. Tìm Min của: \(D=\dfrac{x+y}{\sqrt{xy}}+\dfrac{\sqrt{xy}}{x+y}\)
Làm hộ mình câu c nha
Cho \(H=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x^3}-\sqrt{y^3}}{x-y}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\).
a) Rút gọn H
b) Chứng minh \(H\ge0\)
c) So sánh H với \(\sqrt{H}\)
Cho x,y,z >0. Chứng minh rằng :
\(\dfrac{\sqrt{xy}}{1+\sqrt{yz}}+\dfrac{1}{\sqrt{xy}+\sqrt{yz}}+\sqrt{\dfrac{2\sqrt{yz}}{1+\sqrt{xy}}}\ge2\)
Rút gọn biểu thức:
\(P=\left(\dfrac{1}{xy\sqrt{y}}-\dfrac{1}{xy\sqrt{x}}\right):\left(\dfrac{1}{x^2+xy+2x\sqrt{xy}}+\dfrac{1}{xy+y^2+2y\sqrt{xy}}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
\(D=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-1\right)\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}+1\right)\)
Rút gọn D
Bài 1: Rút gọn
d) \(\dfrac{\sqrt{xy^{2^{ }}}.\sqrt{x^{2^{ }}-y^2}}{\sqrt{\left(x+y\right)\left(x^{2^{ }}y^{3^{ }}-xy^4\right)}}\) ( với x>y>0)
Ta có A=\(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}=\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}}{\sqrt{y}}\)
