a/ĐK:x\(\ge0\);\(x\ne1\)
Ta có: \(P=\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{\sqrt{x^3}-x}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{\left(\sqrt{x-1}-\sqrt{x}\right)\left(\sqrt{x-1}+\sqrt{x}\right)}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\frac{2\sqrt{x-1}}{x-1-x}+x\)
\(=-2\sqrt{x-1}+x\)
Vậy P\(=-2\sqrt{x-1}+x\) khi x\(\ge0\);\(x\ne1\)
b/Để P=9
=>-\(2\sqrt{x-1}+x\)=9
\(\Leftrightarrow\left(x-1\right)-2\sqrt{x-1}+1=9\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=9\)
\(\Leftrightarrow\sqrt{x-1}-1=3\)( do \(\sqrt{x-1}>0\))
\(\sqrt{x-1}=4\Leftrightarrow x-1=16\Leftrightarrow x=17\)
Vậy để P=9 thì x=17
c/Ta có:\(P=x-2\sqrt{x-1}=\left(x-1\right)-2\sqrt{x-1}+1=\left(\sqrt{x-1}-1\right)^2\ge0\)
Chúc bạn học tốt!
