Question

Chứng minh: \(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=x-y\) với \(x>0,y>0\)

Answer 1

Ta có:

\(VT=\left(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\right)\\ =\left(\frac{\sqrt{x}\cdot\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\right)\\ =\left(\frac{\sqrt{xy}\left[\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2\right]}{\sqrt{xy}}\right)\\ =x-y=VP\left(đpcm\right)\)

Vậy với x>0, y>0 ta có đpcm

Answer 2

\(\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)= x-y

=\(\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)

= \(x-y=x-y\)