với a>0;a≠1 thì
\(M=\left(\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-2\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\right)\left(1+\dfrac{1}{a}\right)\) \(=\left(\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right).\dfrac{a+1}{a}\) \(=\left(\dfrac{1-\sqrt{a}+1+\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right).\dfrac{a+1}{a}=\left(\dfrac{2}{2\left(1-a\right)}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right).\dfrac{a+1}{a}=\left(\dfrac{1}{1-a}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right).\dfrac{a+1}{a}=\dfrac{1+a-a^2-1}{\left(1-a\right)\left(1+a\right)}.\dfrac{a+1}{a}=\dfrac{a-a^2}{\left(1-a\right)\left(1+a\right)}.\dfrac{a+1}{a}=\dfrac{a\left(1-a\right)}{\left(1-a\right)\left(1+a\right)}.\dfrac{a+1}{a}=1\) Vậy giá trị của biểu thức M không phụ thuộc vào a
