\(\dfrac{1}{a^3}< \dfrac{1}{\left(a-1\right).a.\left(a+1\right)}\)

\(\Leftrightarrow a^3>a\left(a-1\right)\left(a+1\right)\) vì \(a\inℕ\)

\(\Leftrightarrow a^3>a\left(a^2-1\right)\)

\(\Leftrightarrow a^3>a^3-a\)

\(\Leftrightarrow-a< 0\) (đúng do \(a\inℕ\)) 

Suy ra đpcm. 

mình biết

\(\frac{a}{n\left(n+a\right)}\)

=\(\frac{\left(n+a\right)-n}{n\left(n+a\right)}\)

=\(\frac{n+a}{n\left(n+a\right)}\)\(-\frac{n}{n\left(n+a\right)}\)

Rút gọn, ta được:

\(\frac{1}{n}\)\(-\frac{1}{n+a}\)

=>đpcm

A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

A=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

A=\(\frac{1}{2}-\frac{1}{100}\)

A=\(\frac{50}{100}-\frac{1}{100}\)

A=\(\frac{49}{100}\)

Sai đề bài r phải là như này chứ: CMR \(\frac{a}{n.\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)

Giải: \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n.\left(n+a\right)}=\frac{a}{n.\left(n+a\right)}\)

ta thấy :1/n - 1/n+a=(n+a)-n/n.(n+a)

Ta có:

\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)

Hay  \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\left(đpcm\right)\)

Ta có: \(\frac{1}{n}\).\(\frac{1}{n+1}\)=\(\frac{1}{n.n+1}\) 

       \(\frac{1}{n}\) -\(\frac{1}{n+1}\)=\(\frac{n+1}{n.\left(n+1\right)}\)-\(\frac{n}{n.\left(n+1\right)}\)

                    =\(\frac{n+1-n}{n.\left(n+1\right)}\)=\(\frac{1}{n.\left(n+1\right)}\)

Chọn mình nhé

P = a+a^2+a^3+...+a^2n

P = (a+a^2) + (a^3+a^4)+...+(a2n-1+a2n)

P = a(1+a)+ a^3(1+a)+....+a^2n-1(1+a)

P = (a+1)(a+a^3+...+a^2n-1) chia hết cho a+1 (đpcm)