Đáp án: $P=\dfrac{x+2}{\sqrt{x}(\sqrt{x}-1)}$
Giải thích các bước giải:
ĐKXĐ: $x>0, x\ne 1$
Ta có:
$P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x-1}-\dfrac{2}{\sqrt{x}-x}$
$\to P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}-\dfrac{2}{\sqrt{x}(1-\sqrt{x})}$
$\to P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}+\dfrac{2}{\sqrt{x}(\sqrt{x}-1)}$
$\to P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}(\sqrt{x}-1)}$
$\to P=\dfrac{\sqrt{x}+1-1}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}(\sqrt{x}-1)}$
$\to P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}(\sqrt{x}-1)}$
$\to P=\dfrac{x}{\sqrt{x}(\sqrt{x}-1)}+\dfrac{2}{\sqrt{x}(\sqrt{x}-1)}$
$\to P=\dfrac{x+2}{\sqrt{x}(\sqrt{x}-1)}$
