Question

[NT Ánh]

Rút gọn:

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

b) \(\frac{2}{2a-1}\)\(\sqrt{5a^2\left(1-4a+4a^2\right)}\) với a>0,5

Answer 1

[Trần Việt Linh]

a) \(\frac{2}{x^2-y^2}\cdot\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{\sqrt{3}\left(x+y\right)}{\sqrt{2}}=\frac{\sqrt{6}}{x-y}\)

b) \(\frac{2}{2a-1}\cdot\sqrt{5a^2\left(1-4a+4a^2\right)}=\frac{2}{2a-1}\cdot\sqrt{5a^2\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}\cdot\sqrt{5}a\left(1-2a\right)=-2\sqrt{5}a\)