Hình như sai đề thì phải chứ mk làm ko đc !!!

  A < 1/(1.2) + 1/(2.3) + 1/(3.4) + ...+ 1/(99.100) 
<=> A< 1- 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + .. + 1/99 - 1/100 
<=> A < 1 - 1/100 < 1 (đpcm) 

So với  thì đây

có: 1/3^2<1/2.3; 1/4^2<1/3.4:...: 1/100^2<1/99.100

Mà: 1/1.2+1/2.3+...+1/99.100=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

=> 1/3^2+1/4^2+...+1/100^2<99/100<1

=> đpcm

UNDERSTAND ???

đặt A= biểu thức trên

tao có 

A<1/2.3+1/3.4+...+1/99.100

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

A<1/2-1/100<1/2

SUY RA A<1/2(DPCM)

ai lam day du dau tien minh se k cho nha

minh can gap lam

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

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

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

             ...

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

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

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

\(\Rightarrow A< 1-\frac{1}{100}< 1\)

\(\Rightarrow A< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\).

Đặt \(A=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{100!}\)

Ta thấy:

\(\dfrac{1}{2!}=\dfrac{1}{1.2};\dfrac{1}{3!}=\dfrac{1}{1.2.3}< \dfrac{1}{2.3};...;\dfrac{1}{100!}=\dfrac{1}{1.2...100}< \dfrac{1}{99.100}\)

Cộng vế với vế ta được:

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

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

\(\Rightarrow A< 1-\dfrac{1}{100}< 1\)

Vậy \(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{100!}< 1\) (Đpcm)

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

Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có : 

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

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

\(A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

Vậy \(A< \frac{1}{2}\) ( đpcm ) 

Chúc bạn học tốt ~ 

cam on

Ta có : 1/2 = 0,5

            2/3 = 0,666...

=> 1/2 + 2/3 + ... + 99/100 = 0,5 + 0,666...+3/4 + ... + 99/100

                                           = 1,1,6666... + 3/4 + ... +99/100 > 1

=> 1/2 + 2/3 + ... + 99/100 > 1

 \(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\le1\)

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

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

\(=1-\frac{1}{100}\le1\)

\(\Rightarrow1-\frac{1}{100}\le1\)

1/2 + 2/3 + 3/4 + ... + 99/100 < 1

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

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

= 1 - 1/100 < 1 (đpcm)