Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

Vì \(\frac{49}{50}

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

vi /chia au cong thi cha be hon a

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

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

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

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

~~~

#Sunrise

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

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

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

\(\Rightarrowđpcm\)

1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)

=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)

=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25

=1/26+1/27+...+1/50 (đpcm)

1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)

=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)

=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25

=1/26+1/27+...+1/50 (đpcm)

Hình như sai đề rồi bạn ạ

what ....... what .......what 

1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)

=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)

=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25

=1/26+1/27+...+1/50 (đpcm)

ỦNg hộ nhà mih lại cho !!!

1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)

=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)

=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25

=1/26+1/27+...+1/50 (đpcm)

Ung ho nha mih lai cho

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}< 1\)

Ta có: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}< 1\)

=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}< 1\)

= \(\dfrac{1}{1}-\dfrac{1}{50}< 1\)

= \(\dfrac{50}{50}+\dfrac{-1}{50}< 1\)

= \(\dfrac{49}{50}< 1\)

Vậy \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}< 1\)

1/1.2 = 2 đã lớn hơn 1 rồi @@

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

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

giúp mình nhé

Đặt :

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

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

\(A=1-\dfrac{1}{100}\)

\(A=\dfrac{99}{100}< 1\)

\(\Rightarrow A< 1\)

\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+.........+\dfrac{1}{99.100}< 1\rightarrowđpcm\)

a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)

=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)

=> \(1-\frac{1}{x+1}=\frac{44}{45}\)

=> \(\frac{x}{x+1}=\frac{44}{45}\)

=> x = 44

b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

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

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

\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)

a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45

=> x=44

b/ 1/2² < 1/1.2; 1/3² < 1/2.3; ....; 1/45² < 1/44.45

=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1

=> A < 1