Đáp án:
$f,\sqrt x$
$g,4$
$h,1-a$
Giải thích các bước giải:
f) $\dfrac{x+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\,\,(ĐK:x\ge 0;\,y\ge 0\,x^2+y^2\ne 0)$
$=\dfrac{\sqrt{x}.(\sqrt{x}+\sqrt{y})}{\sqrt{x}+\sqrt{y}}$
$=\sqrt{x}$
g) $\left(\dfrac{\sqrt x+\sqrt y}{\sqrt x-\sqrt y}-\dfrac{\sqrt x-\sqrt y}{\sqrt x+\sqrt y}\right):\dfrac{\sqrt{xy}}{x-y}\,\,(ĐK:\,x\ge 0;\,y\ge 0;\,x\ne y)$
$=\left(\dfrac{(\sqrt x+\sqrt y)^2}{(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)}-\dfrac{(\sqrt x-\sqrt y)^2}{(\sqrt x+\sqrt y)(\sqrt x-\sqrt y)}\right):\dfrac{\sqrt{xy}}{x-y}$
$=\dfrac{(\sqrt x+\sqrt y-\sqrt x+\sqrt y)(\sqrt x+\sqrt y+\sqrt x-\sqrt y)}{x-y}:\dfrac{\sqrt{xy}}{x-y}$
$=\dfrac{2\sqrt y.2\sqrt x}{x-y}.\dfrac{x-y}{\sqrt x.\sqrt y}$
$=4$
h) $\left(1+\dfrac{a+\sqrt a}{\sqrt a+1}\right).\left(1-\dfrac{a-\sqrt a}{\sqrt a-1}\right)\,\,(ĐK:a\ge 0;\,a\ne 1)$
$=\left(1+\dfrac{\sqrt a.(\sqrt a+1)}{\sqrt a+1}\right).\left(1-\dfrac{\sqrt a.(\sqrt a-1)}{\sqrt a-1}\right)$
$=\left(1+\sqrt a\right).\left(1-\sqrt a\right)$
$=1-(\sqrt a)^2$
$=1-a$.
