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

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

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

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

\(B=1.2+2.3+3.4+...+49.50\)

\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(=49.50.51\)

\(B=\frac{49.50.51}{3}=49.50.17\)

\(50^2.A-\frac{B}{17}=49.50-49.50=0\)

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

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

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

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

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

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

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

Sửa đề: 2/2.3 ➜ 1/2.3

Giải:

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

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

M=1/1-1/50

M=49/50

Vì 49/50<1 nên M<1

Chúc bạn học tốt!

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

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

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

bai toan nay kho

Đinh Đức Hùng giải đúng rồi đó

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}\\ =\dfrac{49}{50}\)

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

A= 1- 1/50= 49/50

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

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

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

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

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

=1-1/50<1