Đáp án: $P=\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}$
Giải thích các bước giải:
Ta có:
$P=\left(\dfrac{\sqrt{x}}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}$
$\to P=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}$
$\to P=\dfrac{\sqrt{x}-\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}$
$\to P=\dfrac{-\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}$
$\to P=\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}$
$\to P=\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}$
