a) ĐKXĐ: x≠1;x≥0
Ta có P=(\(\dfrac{1}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\)): \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}\right)^2-2\sqrt{x}+1^2}\)
P=\(\left[\dfrac{1}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\right]\):\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
P=\(\dfrac{\sqrt{x}+1}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)*\(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
P=\(\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
P=\(\dfrac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}{x}\)
Vậy P=với x≠1; x≥0
b) Ta có P=-1⇔\(\dfrac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}{x}=-1\)
⇒\(\sqrt{x}\cdot\left(\sqrt{x}-1\right)=-x\)
⇔x-\(\sqrt{x}=-x\)
⇔\(2x=\sqrt{x}\)
⇔\(2\sqrt{x}=1\)
⇔\(\sqrt{x}=\dfrac{1}{2}\)
⇒\(x=\dfrac{1}{4}\)(thỏa mãn ĐKXĐ)
Vậy khi P=-1 thì \(x=\dfrac{1}{4}\)
