ĐKXĐ: \(x\ge0;x\ne4\)
a.
\(A=\dfrac{\sqrt{9}}{\sqrt{9}-2}=\dfrac{3}{3-2}=3\)
b.
\(T=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}-\dfrac{4\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}-2\sqrt{x}+4-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
c.
\(T=1-\dfrac{4}{\sqrt{x}+2}\ge1-\dfrac{4}{0+2}=-1\)
\(T=1-\dfrac{4}{\sqrt{x}+2}< 1\)
\(\Rightarrow-1\le T< 0\Rightarrow T=\left\{-1;0\right\}\)
Với \(T=-1\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=-1\Rightarrow x=0\)
Với \(T=0\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=0\Rightarrow x=4\)
Vậy \(x=\left\{0;4\right\}\) thì T nguyên
