Question

rút gọn biểu thức

\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\) với x\(\ge\)y; y\(\ge\)0 ; x\(\ne\)y

Answer 1

Ta có : \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2\sqrt{3}.\left|x+y\right|}{\sqrt{2}.\left(x-y\right)\left(x+y\right)}\)

Vì \(x\ge y\ge0\) nên ta có : \(\left|x+y\right|=x+y\)

\(\Rightarrow\frac{2\sqrt{3}\left|x+y\right|}{\sqrt{2}\left(x-y\right)\left(x+y\right)}=\frac{\sqrt{2}.\sqrt{6}\left(x+y\right)}{\sqrt{2}\left(x-y\right)\left(x+y\right)}=\frac{\sqrt{6}}{x-y}\)