Đáp án: $ P=\dfrac{2x+1}{\sqrt{x}}$

Giải thích các bước giải:

Ta có:

$P=(\dfrac{\sqrt{x}+1}{x-\sqrt{x}}+\dfrac{\sqrt{x}+1}{x+2\sqrt{x}+1}-\dfrac1{x-1}):\dfrac1{x-1}$

$\to P=(\dfrac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}+\dfrac{\sqrt{x}+1}{(\sqrt{x}+1)^2}-\dfrac1{(\sqrt{x}+1)(\sqrt{x}-1)})\cdot (x-1)$

$\to P=(\dfrac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}+\dfrac{1}{\sqrt{x}+1}-\dfrac1{(\sqrt{x}+1)(\sqrt{x}-1)})\cdot (x-1)$

$\to P=\dfrac{(\sqrt{x}+1)^2+\sqrt{x}(\sqrt{x}-1)-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-1)}\cdot (x-1)$

$\to P=\dfrac{2x+1}{\sqrt{x}(x-1)}\cdot (x-1)$

$\to P=\dfrac{2x+1}{\sqrt{x}}$