\(dk:x\ge0;x\ne1;P=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}-1}{x-\sqrt{x}+1}=\left(\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{x-\sqrt{x}+1}=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}:\frac{\sqrt{x}-1}{x-\sqrt{x}+1}=\frac{1}{x-\sqrt{x}+1}.\frac{x-\sqrt{x}+1}{\sqrt{x}-1}=\frac{1}{\sqrt{x}-1};P=10\Leftrightarrow\sqrt{x}-1=\frac{1}{10}\Leftrightarrow\sqrt{x}=\frac{11}{10}\Leftrightarrow x=\frac{121}{100}\left(tm\right)\)
P=\(\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}-1}{x-\sqrt{x}+1}\) (đk: \(x\ge0,x\ne1\) )
=\(\left(\frac{x+2}{\sqrt{x^3}+1}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right).\frac{x-\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{x-\sqrt{x}+1}{\sqrt{x}-1}\)
=\(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{1}{\sqrt{x}-1}\)
b, Để P=10
<=> \(\frac{1}{\sqrt{x}-1}=10\)
<=> \(10\sqrt{x}-10=1\) <=> \(10\sqrt{x}=11\) <=> \(\sqrt{x}=\frac{11}{10}\)
=> \(x=\left[{}\begin{matrix}-\sqrt{\frac{10}{11}}\\\sqrt{\frac{10}{11}}\end{matrix}\right.\)(tm)
Vậy P=10 <=> x\(\in\left\{-\sqrt{\frac{11}{10}},\sqrt{\frac{11}{10}}\right\}\)
