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

Kiyotaka Ayanokoji · Accepted Answer

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 $B=\frac{1}{a+1}$Câu trả lời: 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 $B=\frac{1}{a+1}$

๖²⁴ʱんuﾘ ｲú❄✎﹏ · Answer

$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)$$=\frac{1}{a+1}$Vậy $B=\frac{1}{a+1}$Câu trả lời: $B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)ĐK\left(a\ge0;a\ne1\right)$$=\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 $B=\frac{1}{a+1}$

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