Question

Answer 1

làm bài 7+8 thôi nhé

 

Answer 2

Bài 7:

a) \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

b) \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)

Bài 8:

a) \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{19.20}\\ A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\\ A=1-\dfrac{1}{20}\\ A=\dfrac{19}{20}\)

b) \(B=\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{132}\\ B=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{11.12}\\ B=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{11}-\dfrac{1}{12}\\ B=\dfrac{1}{4}-\dfrac{1}{12}\\ B=\dfrac{1}{6}\)

Answer 3

Answer 4

Bài 8:

a: Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{19\cdot20}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)

b: Ta có: \(B=\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{132}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{11}-\dfrac{1}{12}\)

\(=\dfrac{1}{4}-\dfrac{1}{12}\)

\(=\dfrac{3}{12}-\dfrac{1}{12}=\dfrac{1}{6}\)

Answer 5

Bài 7:

a: \(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}=\dfrac{1}{2\cdot3}\)

b: \(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)