Question

Tính giá trị biểu thức sau:
1.\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)......\left(1-\dfrac{1}{n+1}\right)\)với n thuộc N
2.\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

3.\(C=-66.\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

4.\(D=\dfrac{7}{4}\left(\dfrac{33}{12}\dfrac{3333}{2020}\dfrac{333333}{303030}\dfrac{33333333}{42424242}\right)\)

giúp mik nhé

Answer 1

1, \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...........\left(1-\dfrac{1}{n+1}\right)\)

\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)...........\left(\dfrac{n+1}{n+1}-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}..............\dfrac{n}{n+1}\)

\(=\dfrac{1.2.3........n}{2.3.......\left(n+1\right)}\)

\(=\dfrac{1}{n+1}\)

2, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

Answer 2

C=\(-66\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

=\(-66.\left(\dfrac{5}{66}\right)+124\left(-37-63\right)=-5+124.\left(-100\right)\)

=-12405

Answer 3

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{n+1}\right)\)

=\(\dfrac{1}{2}\).\(\dfrac{2}{3}\).\(\dfrac{3}{4}\).....\(\dfrac{n}{n+1}\)=\(\dfrac{1}{n+1}\)

Vậy...

Answer 4

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Answer 5

\(D=\dfrac{7}{4}\left(\dfrac{33}{12}.\left(\dfrac{3333}{2020}\right).\left(\dfrac{333333}{303030}\right).\left(\dfrac{33333333}{42424242}\right)\right)\)

=\(\dfrac{7}{4}.\left(\dfrac{3.11}{2.6}.\left(\dfrac{3.1111}{2.1010}\right)\left(\dfrac{3.111111}{2.151515}\right)\left(\dfrac{3.11111111}{2.21212121}\right)\right)\)

=\(\dfrac{7}{4}.\left(\dfrac{3}{2}\right)\left(\dfrac{11}{6}\right)\left(\dfrac{1111}{1010}\right)\left(\dfrac{111111}{151515}\right)\left(\dfrac{11111111}{21212121}\right)\)

Chịu ==''