\(\dfrac{3.3}{20.23}+\dfrac{3.3}{23.26}+...+\dfrac{3.3}{77.80}\)

\(=3\left(\dfrac{3}{20.23}+\dfrac{3}{23.26}+...+\dfrac{3}{77.80}\right)\)

\(=3\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)\)

\(=3\left(\dfrac{1}{20}-\dfrac{1}{80}\right)\)

\(=3.\dfrac{3}{80}=\dfrac{9}{80}< 1\left(đpcm\right)\)

Vậy...

rất là hay

1/2*2+1/3*3+...+1/2017*2017

Đáp số 2016

3x3 =9

3.3=nine.

Trả lời ;>..........................................................

2

7

9

Hk otot,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k nhé

1/2.2 < 1/1.2

1/3.3 < 1/2.3

..................

1/100.100 < 1/99.100 

=> <

Ta có: \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\)

Vì \(\frac{1}{2^2}<\frac{1}{1.2}\)

\(\frac{1}{3^2}<\frac{1}{2.3}\)

\(\frac{1}{4^2}<\frac{1}{3.4}\)

.....

\(\frac{1}{100^2}<\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}<1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<1\left(đpcm\right)\)

1/2.2 < 1/1.2

1/3.3 < 1/2.3

..................

1/100.100 < 1/99.100 

=> <

\(< \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\left(đpcm\right)\)

sai đề

1/3.3+1/4.4+1/5.5+...+1/100.100<1/2.3+1/3.4+1/4.5+...+1/99.100=A

A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100=1/2-1/100<1/2

=>1/3.3+1/4.4+1/5.5+...+1/100.100<1/2

=\(\dfrac{7}{2}\)