Question

giúp mình với

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

 

Answer 1

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

1-1/2+1/2-1/3+/13-1/4+1/4-1/5+1/5-...-1/49+1/49-1/50

1-1/50

50/50-1/50=49/50

Answer 2

E=1/1*2+1/2*3+1/3*4+...+1/49*50

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

E=1-1/50

E=49/50

 

Answer 3

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

Answer 4

Ta có: \(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

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

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

Vậy E=\(\frac{49}{50}\)

Answer 5

may cau nay de lam ban

Answer 6

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

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

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

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

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

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