qui đồng ps ta dc

1/n-1/n+1=n+1-n/n(n+1)=1/n(n+1)

nhanh tay

Xét vế trái, ta có :

   1 / n       -      1 / ( n + 1 )

= ( n + 1 ) / n( n + 1 )    -     n / n( n + 1 )

= n + 1 - n / n( n + 1 )

= 1 / n( n + 1 ) ( Vế phải )

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

\(=\dfrac{n+2}{2n\left(n+1\right)\left(n+2\right)}-\dfrac{n}{2n\left(n+1\right)\left(n+2\right)}=\dfrac{1}{2}\left[\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right]\)

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

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

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

Ta có: \(n\left(n-1\right)=n^2-n< n^2\Rightarrow\dfrac{1}{n\left(n-1\right)}>\dfrac{1}{n^2}\)

\(n\left(n+1\right)=n^2+n>n^2\Rightarrow\dfrac{1}{n\left(n+1\right)}< \dfrac{1}{n^2}\)

Từ đó:

\(\dfrac{1}{n-1}-\dfrac{1}{n}=\dfrac{n-\left(n-1\right)}{n\left(n-1\right)}=\dfrac{1}{n\left(n-1\right)}>\dfrac{1}{n^2}\) (1)

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

(1);(2) \(\Rightarrow\dfrac{1}{n-1}-\dfrac{1}{n}>\dfrac{1}{n^2}>\dfrac{1}{n}-\dfrac{1}{n+1}\) (đpcm)

ta có : 1/n - 1/ n+1 =n+1/n.(n+1) - n/n(n+1)

                              =1/n(n+1)

Vậy ta có đpcm

\(\frac{1}{n}-\frac{1}{n}+1=0+1=1\)    (1)

\(\frac{1}{n}.\left(n+1\right)=\frac{1}{n}.n+\frac{1}{n}.1=1+\frac{1}{n}\)        (2)

Vì n là mẫu nên n\(\ne\)0. Vậy từ (1) và (2) suy ra không chứng minh được.

Bạn xem lại đề bài!

\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\)

\(=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

1+2+3+4+5+6+7+8+9=133456 hi hi

đào xuân anh sao mày gi sai hả

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