Question

Rút gọn và tính

Answer 1

\(\left(\dfrac{y\sqrt{y}-x\sqrt{x}}{\sqrt{y}-\sqrt{x}}+\sqrt{xy}\right)\left(\dfrac{\sqrt{y}-\sqrt{x}}{y-x}\right)^2\)

\(=\left(\sqrt{xy}+\sqrt{xy}\right)\left(\sqrt{y}+\sqrt{x}\right)^2=2\sqrt{xy}\left(x+2\sqrt{xy}+y\right)\)

\(=2x\sqrt{xy}+4xy+2y\sqrt{xy}\)

Answer 2

sửa bài : \(\left(\dfrac{y\sqrt{y}-x\sqrt{x}}{\sqrt{y}-\sqrt{x}}+\sqrt{xy}\right)\left(\dfrac{\sqrt{y}-\sqrt{x}}{y-x}\right)^2\)ĐK : \(x\ne y;x;y>0\)

\(=2\sqrt{xy}\left(\dfrac{1}{\sqrt{y}+\sqrt{x}}\right)^2=\dfrac{2\sqrt{xy}}{x+2\sqrt{xy}+y}\)

Answer 3

Ta có: \(\left(\dfrac{y\sqrt{y}-x\sqrt{x}}{\sqrt{y}-\sqrt{x}}+\sqrt{xy}\right)\cdot\left(\dfrac{\sqrt{y}-\sqrt{x}}{y-x}\right)^2\)

\(=\left(y+2\sqrt{xy}+x\right)\cdot\dfrac{1}{\left(\sqrt{y}+\sqrt{x}\right)^2}\)

=1