Question

[NT Dũng]

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

[Nguyễn Thanh Hằng]

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

[Nguyễn Thị Hồng Nhung]

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