Giải thích các bước giải:
a.Ta có:
$\dfrac{3x}{y}\sqrt{\dfrac{y}{x}}=\sqrt{(\dfrac{3x}{y})^2\cdot \dfrac{y}{x}}=\sqrt{\dfrac{9x}{y}}$
b.Ta có:
$(x-2)\sqrt{\dfrac{1}{x-2}}=\sqrt{(x-2)^2\cdot\dfrac{1}{x-2}}=\sqrt{x-2}$
c.ĐKXĐ: $x>\dfrac13$
Ta có:
$(3x-1)\sqrt{\dfrac{1}{3x-1}}=\sqrt{(3x-1)^2\cdot \dfrac{1}{3x-1}}=\sqrt{3x-1}$
d.Ta có:
$\dfrac{x+2\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\cdot \sqrt{\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}}$
$=\dfrac{(\sqrt{x}+\sqrt{y})^2}{\sqrt{x}+\sqrt{y}}\cdot \sqrt{\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}}$
$=(\sqrt{x}+\sqrt{y})\cdot \sqrt{\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}}$
$=\sqrt{(\sqrt{x}+\sqrt{y})^2\cdot \dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}}$
$=\sqrt{(\sqrt{x}+\sqrt{y})\cdot(\sqrt{x}-\sqrt{y})}$
$=\sqrt{x-y}$
