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

=1-1/50<1

\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{49.50}\)

\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)

\(M=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+........+\left(-\frac{1}{49}+\frac{1}{49}\right)-\frac{1}{50}\)

\(M=\frac{1}{1}-0+0+0+0+0+......+0+0-\frac{1}{50}\)

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

Vì \(\frac{49}{50}<1\) nên  \(S<1\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

     \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

     \(=1-\frac{1}{50}<1\)

\(\Rightarrow M<1\) 

Vậy \(M<1\)

Chúc bạn học tốt!!!!!!!

M=1/1.2+1/2.3+1/3.4+...+1/49.50

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

M=1-1/50<1

Vậy M<1

M=1/1.2+1/2.3+...+1/49.50

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

M=1-1/50<1

=>M<1

\(M=\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\)

\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(M=1-\frac{1}{50}<1\)

\(=>M<1\)

M = 1/1.2 + 1/2.3 + ... + 1/49.50

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

M = 1 - 1/50

M = 49/50 

Vì 49/50 < 1

=> M < 1

Đề sai! Đề đúng : M = 1/1.2 + 1/2.3 +........+ 1/99.100

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

M = 1 - 1/100

M = 99/100 < 1

=> M < 1

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(M=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{49}+\frac{1}{49}\right)-\frac{1}{50}\)

\(M=1+0+0+...+0-\frac{1}{50}\)

\(M=\frac{49}{50}\)

\(\Rightarrow\frac{49}{50}< 1\)

\(\Rightarrow M< 1\)

dấu chấm ở giữa hai số là dấu nhân à?

ừ dấu chấm là dấu nhân

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(M=\frac{1}{1}+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}\)

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

\(M=\frac{49}{50}\)

Vậy M < 1

\(M=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{49}-\frac{1}{50}\)

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

Vì \(\frac{49}{50}< 1\)\(\Rightarrow M< 1\)

VẬY M < 1

HK TỐT #

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

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

\(\Leftrightarrow M< 1\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)

      \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

     \(=1-\frac{1}{50}\)

Mà \(1-\frac{1}{50}< 1\)nên \(M< 1\)

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

= \(1-\frac{1}{50}

Ta có : 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 

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

= 1 - 1/50 < 1

Nên 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 < 1

Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}+\dfrac{1}{2020\cdot2021}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(=\dfrac{1}{1}-\dfrac{1}{2021}=\dfrac{2021}{2021}-\dfrac{1}{2021}\)

\(=\dfrac{2020}{2021}\)

mà \(\dfrac{2020}{2021}< \dfrac{2021}{2021}=1\)

nên A<1