Ta có: \(M=\left(\dfrac{1}{2\sqrt{a}+2}-\dfrac{1}{2\sqrt{a}-2}+\dfrac{a^2+1}{a^2-1}\right)\left(1+\dfrac{1}{a}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+1\right)}{2\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)\left(a+1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a+1\right)}{2\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\left(a+1\right)}+\dfrac{2a^2+2}{2\left(a-1\right)\left(a+1\right)}\right)\cdot\dfrac{a+1}{a}\)
\(=\dfrac{a\sqrt{a}+\sqrt{a}-a-1-\left(a\sqrt{a}+\sqrt{a}+a+1\right)+2a^2+2}{2\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)\left(a+1\right)}\cdot\dfrac{a+1}{a}\)
\(=\dfrac{a\sqrt{a}+\sqrt{a}-a-1-a\sqrt{a}-\sqrt{a}-a-1+2a^2+2}{2\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{1}{a}\)
\(=\dfrac{2a^2-2a}{2\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{1}{a}\)
\(=\dfrac{2a\left(a-1\right)}{2\left(a-1\right)\cdot a}\)
=1
