\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
vậy...
k mình nha
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\left(dpcm\right)\)
\(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(⋮\)
\(\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}\)
= \(\frac{99}{100}\)
Mà : \(\frac{99}{100}< 1\)
=> \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 1\left(đpcm\right)\)
Ta co : 1 . 1/2 + 1/2 . 1/3 + 1/3 . 1/4 + ... + 1/99 . 1/100
= 1 . 1/100
= 1/100
Vi 1/100 nho hon 1
=> 1/1.2 + 1/2.3 + ... + 1/99.100 < 1
MINH lam con k chac chan lam dau
Ta có :\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vì \(\frac{99}{100}< \frac{100}{100}\)nên \(\frac{99}{100}< 1\)
\(\Rightarrow\)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 1\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\Rightarrow\frac{99}{100}< 1\left(đpcm\right)\)
