Bài 11:

a: \(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)

b: A>0

=>\(\frac{\sqrt{a}-2}{3\sqrt{a}}>0\)

=>\(\sqrt{a}-2>0\)

=>\(\sqrt{a}>2\)

=>a>4

c: Đặt B=1:A

\(=1:\frac{\sqrt{a}-2}{3\sqrt{a}}=\frac{3\sqrt{a}}{\sqrt{a}-2}\)

Để B là số nguyên thì \(3\sqrt{a}\) ⋮\(\sqrt{a}-2\)

=>\(3\sqrt{a}-6+6\) ⋮\(\sqrt{a}-2\)

=>6⋮\(\sqrt{a}-2\)

=>\(\sqrt{a}-2\in\left\lbrace1;-1;2;-2;3;-3;6;-6\right\rbrace\)

=>\(\sqrt{a}\) ∈{3;1;4;0;5;-1;8;-4}

=>\(\sqrt{a}\) ∈{0;1;3;4;5;8}

=>a∈{0;1;9;16;25;64}

Kết hợp ĐKXĐ, ta được: a∈{9;16;25;64}

Bài 10:

a: \(A=\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{\sqrt{x}+3}{\sqrt{x}-3}-\frac{12\sqrt{x}}{x-9}\)

\(=\frac{\left(\sqrt{x}-3\right)^2+\left(\sqrt{x}+3\right)^2-12\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x-6\sqrt{x}+9+x+6\sqrt{x}+9-12\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{2x-12\sqrt{x}+18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)

b: A=1

=>\(2\left(\sqrt{x}-3\right)=\sqrt{x}+3\)

=>\(2\sqrt{x}-6=\sqrt{x}+3\)

=>\(\sqrt{x}=9\)

=>x=81(nhận)

d: Để A là số nguyên thì \(2\left(\sqrt{x}-3\right)\vdots\sqrt{x}+3\)

=>\(2\sqrt{x}-6\vdots\sqrt{x}+3\)

=>\(2\sqrt{x}+6-12\vdots\sqrt{x}+3\)

=>\(-12\vdots\sqrt{x}+3\)

=>\(\sqrt{x}+3\in\left\lbrace3;4;6;12\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;1;3;9\right\rbrace\)

=>x∈{0;1;9;81}

Kết hợp ĐKXĐ, ta được: x∈{0;1;81}