Gọi \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

           \(A=1-\frac{1}{100}\)(TỐI GIẢN CÁC PHÂN SỐ LẬP LẠI )

           \(A=\frac{99}{100}

Ta có \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
        = \(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
        = \(\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{100}{99.100}-\frac{99}{99.100}\)
        =\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
        =   \(1-\frac{1}{100}\)
        =     \(\frac{99}{100}\)
Vậy\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)

   1/1.2 + 1/2.3 + 1/3.4 + .......................+ 1/99.100    

= 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 +..................+ 1/99 - 1/100 

= 1 - 1/100 

= 99/100 

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

Ma 99/100 < 1.

=> 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 < 1 (dccm)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

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 : \(\frac{99}{100}< 1\)

=>  \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 1\left(đpcm\right)\)

a) 1/1.2 + 1/2.3 + 1/3.4 + ....... + 1/99.100

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/99 - 1/100

= 1 - 1/100

= 99/100 < 1 nên 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100 < 1 (ĐPCM)

a)1-1/2+1/2-1/3+1/3-1/4+......+1/99-1/100

1-1/100=99/100<1

cho mk nha ^^

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

Vì \(\frac{1}{100}>0\Rightarrow1-\frac{1}{100}< 1\)

\(\frac{\Rightarrow1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}< 1\)

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

\(=1-\frac{1}{100}< 1\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(=\frac{99}{100}< 1\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(=\frac{99}{100}< 1\)

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

\(=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)

\(=1+0+0+...+0-\frac{1}{100}\)

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

\(=\frac{99}{100}< 11\)

Vậy : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 11\)

=1/2-1/3+1/3-1/4+...+1/99-1/100

=1/2-1/100

=50/100-1/100

=49/100<1

=> dãy trên < 1 đđcm

1 chứ ko phai 11 đâu nha

a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

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

\(=1-\dfrac{1}{100}< 1\).

b. Có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\).

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