Bài 1:
Ta có:
\(\frac{a}{b+1}+\frac{-a}{b}=\frac{a}{b+1}-\frac{a}{b}=\frac{ab-a\left(b+1\right)}{\left(b+1\right)b}=\frac{ab-ab-a}{b^2+b}=\frac{-a}{b^2+b}\left(đpcm\right)\)
Bài 2:
Ta có:
\(a^2\ge0\Rightarrow a^2+2015>0\)
⇒Để M>0 thì \(a-2014>0\Rightarrow a>2014\)
Vậy để M=\(\left(a^2+2015\right)\left(a-2014\right)>0\) thì a>2014