Question

[mạc trần]

Rút gọn B=\(\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\cdot\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)\)

Answer 1

[Kiyotaka Ayanokoji]

Trả lời 

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)\)  \(\left(a\ge0.a\ne1\right)\)

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{1}{\left(a+1\right)^2}-\frac{1}{\left(a-1\right).\left(a+1\right)}\right]\)

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{a-1-a-1}{\left(a+1\right)^2.\left(a-1\right)}\right]\)

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.0\)

\(B=\frac{1}{a+1}\)

Vậy \(B=\frac{1}{a+1}\)

Answer 2

[๖²⁴ʱんuﾘ ｲú❄✎﹏]

\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)ĐK\left(a\ge0;a\ne1\right)\)

\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}-\frac{a^2+1}{\left(a^2-1\right)\left(a^2+1\right)}\right)\)

\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1-a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}\right)\)

\(=\frac{1}{a+1}\)

Vậy \(B=\frac{1}{a+1}\)