Question

Rút gọn biểu thức sau :\(\left(\frac{1}{2+2\sqrt{a}}+\frac{1}{2-2\sqrt{a}}-\frac{a^2+1}{1-a^2}\right)\cdot\left(1+\frac{1}{a}\right)\)

Answer 1

Lời giải:

ĐKXĐ: \(a>0; a\neq 1\)

\(\left(\frac{1}{2+2\sqrt{a}}+\frac{1}{2-2\sqrt{a}}-\frac{a^2+1}{1-a^2}\right).\left(1+\frac{1}{a}\right)\)

\(=\left(\frac{1-\sqrt{a}}{2(1+\sqrt{a})(1-\sqrt{a})}+\frac{1+\sqrt{a}}{2(1-\sqrt{a})(1+\sqrt{a})}-\frac{a^2+1}{(1-a)(1+a)}\right).\frac{a+1}{a}\)

\(=\left(\frac{1}{1-a}-\frac{a^2+1}{(1-a)(1+a)}\right).\frac{a+1}{a}\)

\(=\frac{1+a-(a^2+1)}{(1-a)(1+a)}.\frac{a+1}{a}=\frac{a(1-a)}{(1-a)(1+a)}.\frac{a+1}{a}=1\)