a: \(A=\dfrac{2}{\sqrt{x}-2}:\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+4}{x-4}\right)\)
\(=\dfrac{2}{\sqrt{x}-2}:\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\dfrac{2}{\sqrt{x}-2}:\dfrac{3\left(\sqrt{x}+2\right)-\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2}{\sqrt{x}-2}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{3\sqrt{x}+6-\sqrt{x}-4}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)}{2\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
b: Để A là số nguyên thì \(\sqrt{x}+2⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1+1⋮\sqrt{x}+1\)
=>\(1⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;-1\right\}\)
=>\(\sqrt{x}\in\left\{0;-2\right\}\)
=>\(\sqrt{x}=0\)
=>x=0(nhận)
c: \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\left|2+\sqrt{2}\right|-\left|\sqrt{2}+1\right|\)
\(=2+\sqrt{2}-\sqrt{2}-1=1\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+2}{1+1}=\dfrac{3}{2}\)
d: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}\)
\(\sqrt{x}+1>=1\forall x\) thỏa mãn ĐKXĐ
=>\(\dfrac{1}{\sqrt{x}+1}< =\dfrac{1}{1}=1\forall x\) thỏa mãn ĐKXĐ
=>\(\dfrac{1}{\sqrt{x}+1}+1< =2\forall x\) thỏa mãn ĐKXĐ
=>\(A< =2\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
