Question

Rút gọn : \(\frac{a^2}{a^2-1}-\frac{a^2}{1+a^2}.\left(\frac{a}{a+1}+\frac{1}{a^2+a}\right)\)

a) Tìm tập xác định và rút gọn A

b) Tìm a để A = 3

Answer 1

a: ĐKXĐ: \(a\notin\left\{0;1;-1\right\}\)

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

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

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

b: Để A=3 thì \(3a^2-3=a\)

\(\Leftrightarrow2a^2=3\)

hay \(a\in\left\{\dfrac{\sqrt{6}}{2};-\dfrac{\sqrt{6}}{2}\right\}\)