Question

cho bieu thuc A=\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}\)

hay chung to rang 5>A

giup mk nha

Answer 1

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}\)

\(A=\frac{5}{1}-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+...+\frac{5}{99}-\frac{5}{100}=\frac{5}{1}-\frac{5}{100}=\frac{99}{20}<\frac{100.}{20}=5\)

=>A<5 hay 5>A(đpcm)
 

Answer 2

Ta có

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}=5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5.\left(1-\frac{1}{100}\right)\)

Vì \(1-\frac{1}{100}<1\)

=>\(5\left(1-\frac{1}{100}\right)<5.1=5\)

Vậy A<5