\(\dfrac{N}{2}=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}< 1\\ N< 2\)

Ta có: \(N=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{49\cdot50}\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\right)\)

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

\(=2\left(1-\dfrac{1}{50}\right)\)

\(=2\cdot\dfrac{49}{50}=\dfrac{49}{25}< \dfrac{50}{25}=2\)

Vậy: N<2

Ta có: =2(11⋅2+12⋅3+13⋅4+...+149⋅50)=2(11⋅2+12⋅3+13⋅4+...+149⋅50)

=2(1−150)=2(1−150)

Ta có A = 1 / 2 . ( 1 - 1 / 2 + 1 / 2 - 1/ 3 + ............+ 1 / 49 - 1 / 50 )

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

=1/2 + 0 + 0 + .................+ 0 - 1/50

= 1/2 - 1/50

=12/25

Vậy A = 12/25

Ta có 12/25 < 1/2

vậy 25/12 < 1/2

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

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

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

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

A = 49/50

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

A = 1 - 1/50   
A=49/50

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

A = 1/1 - 1/50

A = 49/50

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

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

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

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

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

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

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

`A=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=49/50`

Giải:

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

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

\(A=1-\dfrac{1}{50}\) 

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

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

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

A = 1 - \(\dfrac{1}{50}\)

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

A = \(\dfrac{49}{50}\)

\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)

\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)

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

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

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

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

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

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

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

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

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

\(=\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+.....+\left(-\dfrac{1}{49}+\dfrac{1}{49}\right)+\left(1-\dfrac{1}{50}\right).\)\(=0+0+0+.....+0+\left(1-\dfrac{1}{50}\right).\)

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

=\(\dfrac{49}{50}.\)

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