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

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

m=1-1/2+1/2-1/3+....+1/40-1/50

   = 1/-1/50

   = 49/50

Đặt A=1/1.2+ 1/2.3+1/3.4+1/4.5+...+1/49.50 

=2-1/1.2 + 3-2/2.3+4-3/3.4+5-4/4.5+.......+50-49/49... 

=2/1.2 - 1/1.2+ 3/2.3 - 2/2.3+4/3.4 - 3/3.4 + ................+50/49.50 - 49/49.50 

=1- 1/50 

=49/50 

Vậy giá trị của biểu thức A là 49/50

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

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

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

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

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

\(\Rightarrow M< 1\)

dấu chấm ở giữa hai số là dấu nhân à?

ừ dấu chấm là dấu nhân

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

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

= \(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

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

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

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

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 \(M< 1\)