Đáp án:
$A=1$
$B=\dfrac{x-1}{x}$
Giải thích các bước giải:
$A=\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}$
$=\sqrt{5}-\left(\sqrt{5}-1\right)$
$=\sqrt{5}-\sqrt{5}+1$
$A=1$
$B=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}$ `;(x>0;x!=1)`
$=\dfrac{\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right)\left(x-2\sqrt{x}+1\right)}{\sqrt{x}}$
$=\dfrac{\dfrac{x-1}{\sqrt{x}}}{\sqrt{x}}$
$=\dfrac{x-1}{\sqrt{x}\sqrt{x}}$
$B=\dfrac{x-1}{x}$
