\(\left ( \frac{1}{2+2\sqrt{a}}+\frac{1}{2-2\sqrt{a}}-\frac{a^{2}+1}{1-a^{2}} \right )\left ( 1+\frac{1}{a} \right )\)\(=\left ( \frac{1}{2\left ( 1+\sqrt{a} \right )}+\frac{1}{2\left ( 1-\sqrt{a} \right )}-\frac{a^{2}+1}{1-a^{2}} \right ).\frac{a+1}{a}\)\(=\left ( \frac{ 1-\sqrt{a}+ 1+\sqrt{a}}{2\left ( 1-a \right )}-\frac{a^{2}+1}{1-a^{2}} \right ).\frac{a+1}{a}\)\(=\frac{1+a-a^{2}-1}{1-a^{2}}.\frac{a+1}{a}=\frac{a\left ( a+1 \right )}{\left ( 1-a \right )\left ( 1+a \right )}.\frac{a+1}{a}=1\)
Điều kiện: $a\neq 0.$Khi đó ta có:\(\left ( \frac{1}{2+2\sqrt{a}}+\frac{1}{2-2\sqrt{a}}-\frac{a^{2}+1}{1-a^{2}} \right )\left ( 1+\frac{1}{a} \right )\)
\(=\left ( \frac{1}{2\left ( 1+\sqrt{a} \right )}+\frac{1}{2\left ( 1-\sqrt{a} \right )}-\frac{a^{2}+1}{1-a^{2}} \right ).\frac{a+1}{a}\)
\(=\left ( \frac{ 1-\sqrt{a}+ 1+\sqrt{a}}{2\left ( 1-a \right )}-\frac{a^{2}+1}{1-a^{2}} \right ).\frac{a+1}{a}\)
\(=\frac{1+a-a^{2}-1}{1-a^{2}}.\frac{a+1}{a}=\frac{a\left ( a+1 \right )}{\left ( 1-a \right )\left ( 1+a \right )}.\frac{a+1}{a}=1\)
Vậy biểu thức ban đầu có giá trị bằng 1.