\(\Leftrightarrow2\sqrt{x}-4⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}-3\in\left\{1;-1\right\}\)

hay \(x\in\left\{4;1\right\}\)

a: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}-1}{3-\sqrt{a}}\)

\(=\dfrac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{2\sqrt{a}-9-a+9+2a-5\sqrt{a}+2}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}-3\right)}\)

\(=\dfrac{a-3\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{\sqrt{a}-1}{\sqrt{a}-3}\)

b: A là số nguyên

=>\(\sqrt{a}-3+2⋮\sqrt{a}-3\)

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

=>a thuộc {16;25;1}

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

\(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\dfrac{\sqrt{x}+1-2}{x-1}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

b: Để A là số nguyên thì \(\sqrt{x}-1⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1-2⋮\sqrt{x}+1\)

=>căn x+1 thuộc {1;2}

=>căn x thuộc {0;1}

mà x<>1

nên x=0