Bài 10:
a: ĐKXĐ: x>0; x<>4
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right)\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{2}{\sqrt{x}+2}\)
b: \(B=\frac73\cdot A=\frac73\cdot\frac{2}{\sqrt{x}+2}=\frac{14}{3\sqrt{x}+6}\)
Để B là số nguyên thì 14⋮\(3\sqrt{x}+6\)
=>\(3\sqrt{x}+6\in\left\lbrace7;14\right\rbrace\)
=>\(3\sqrt{x}\in\left\lbrace1;8\right\rbrace\)
=>\(\sqrt{x}\in\left\lbrace\frac13;\frac83\right\rbrace\)
=>x∈{1/9;64/9}
Bài 9:
a: \(P=2\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x-1}+1}\right):\frac{\sqrt{x-1}}{x+\sqrt{x-1}-1}\)
\(=2\cdot\frac{\sqrt{x-1}+1-\sqrt{x-1}}{\sqrt{x-1}\left(\sqrt{x-1}+1\right)}\cdot\frac{\sqrt{x-1}+\sqrt{\left(x-1\right)^2}}{\sqrt{x-1}}\)
\(=2\cdot\frac{1}{\sqrt{x-1}}\cdot\frac{\sqrt{x-1}\left(\sqrt{x-1}+1\right)}{\sqrt{x-1}\left(\sqrt{x-1}+1\right)}=\frac{2}{\sqrt{x-1}}\)
b: Để P là số nguyên thì 2⋮\(\sqrt{x-1}\)
=>\(\sqrt{x-1}\in\left\lbrace1;2\right\rbrace\)
=>x-1∈{1;4}
=>x∈{2;5}
