Đáp án:
$\dfrac{\sqrt{x}+2}{\sqrt{x}-1}$
Giải thích các bước giải:
ĐKXĐ: $x≥0;x\ne1$
$\dfrac{x-1}{x-2\sqrt{x}+1} +\dfrac{\sqrt{x}+1}{x-1}$
$=\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x}-1)^2} +\dfrac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{\sqrt{x}+1}{\sqrt{x}-1} +\dfrac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{(\sqrt{x}+1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)} +\dfrac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{x+2\sqrt{x}+1+\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{x+3\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}$
