Bài 2:
a: \(P=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right)\cdot\frac{\sqrt{x}+2}{\sqrt{x}}\)
\(=\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+2}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
b: P=2
=>\(\frac{2}{\sqrt{x}-2}=2\)
=>\(\sqrt{x}-2=1\)
=>\(\sqrt{x}=3\)
=>x=9(nhận)
Bài 1:
a: \(P=\left(\frac{1}{\sqrt{x}+3}+\frac{2\sqrt{x}}{x-9}\right)\cdot\frac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(=\left(\frac{1}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\frac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}-3+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(=\frac{3\left(\sqrt{x}-1\right)\cdot2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{6}{\sqrt{x}-3}\)
b: P=2
=>\(\frac{6}{\sqrt{x}-3}=2\)
=>\(\sqrt{x}-3=3\)
=>\(\sqrt{x}=3+3=6\)
=>x=36(nhận)
