Question

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

chứng tỏ rằng 5>A

giup mk nha cac bn

Answer 1

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

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

= \(5.\left(1-\frac{1}{100}\right)\)

= \(5-\frac{5}{100}<5\)

=> A < 5 (Đpcm).

Answer 2

cho minh ket ban voi

Answer 3

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

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

A=\(5\left(1-\frac{1}{100}\right)\)

Ta có: 1-\(\frac{1}{100}\)<1

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

=>A<5      (đpcm)