Question

\(A=\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\)

Rút gọn 

Answer 1

Với `a >= 0,a \ne 1` có:

`A=1/[2(1+\sqrt{a})]+1/[2(1-\sqrt{a})]-[a^2+1]/[1-a^2]`

`A=1/[2(1+\sqrt{a})]+1/[2(1-\sqrt{a})]-[a^2+1]/[(1-\sqrt{a})(1+\sqrt{a})]`

`A=[1-\sqrt{a}+1+\sqrt{a}-2a^2-2]/[2(1-\sqrt{a})(1+\sqrt{a})]`

`A=[-2a^2]/[2(1-\sqrt{a})(1+\sqrt{a})]`

`A=[-a^2]/[1-a]`

Answer 2

Lời giải:
ĐKXĐ: $a\geq 0; a\neq 1$

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

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