a) Ta có : ĐKXĐ :\(a\ne4;a\ge0\)

\(\dfrac{\sqrt{a}+2}{\sqrt{a}-2}=\dfrac{\sqrt{a}-2+4}{\sqrt{a}-2}=\dfrac{\sqrt{a}-2}{\sqrt{a}-2}+\dfrac{4}{\sqrt{a}-2}=1+\dfrac{4}{\sqrt{a}-2}\)

Mà \(A\in Z\) => \(1+\dfrac{4}{\sqrt{a}-2}\in Z=>\dfrac{4}{\sqrt{a}-2}\in Z=>\sqrt{a}-2\inƯ\left(4\right)\)

=>\(\left[{}\begin{matrix}\sqrt{a}-2=1\\\sqrt{a}-2=-1\\\sqrt{a}-2=4\\\sqrt{a}-2=-4\end{matrix}\right.< =>\left[{}\begin{matrix}a=9\\a=1\\a=36\\Loại\end{matrix}\right.\)

Ta có :

\(\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}\)

=\(\sqrt{x^2-2.3.x+3^2+4}-\sqrt{x^2-2.3.x+3^2+1}\)

=\(\sqrt{\left(x-3\right)^2+2^2}-\sqrt{\left(x-3\right)^2+1^2}\)

Ta có :

\(\sqrt{4x^2-4x\left(x+1\right)+\left(x+1\right)^2+9}\)

=\(\sqrt{\left(2x\right)^2-2.2x\left(x+1\right)+\left(x+1\right)^2+9}\)

\(=\sqrt{\left(2x-\left(x+1\right)\right)^2+3^2}=\sqrt{\left(x-1\right)^2+3^2}\)

=\(\sqrt{\left(x-1\right)^2}+3\ge3=>Min_A=3\) khi x-1=0=>x=1

p/s Câu b sai đề nha mình chỉnh lại rồi

a) Ta có :

\(\dfrac{1}{\sqrt{x^2-2x-1}}\) xác định khi \(x^2-2x-1=x^2-2x+1-3=\left(x-1\right)^2-1\)

=>\(\left(x-1\right)^2>1< =>x-1>1=>x>2\)

b) c/m tt x>2