Question

tinh gia tri bieu thuc [1+1/100] * [1+1/99]*.....* [1+1/3] * [1+1/2]

Answer 1

Ta có : \(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right).......\left(1+\frac{1}{3}\right)\left(1+\frac{1}{2}\right)\)

\(=\frac{101}{100}.\frac{100}{99}.\frac{99}{98}......\frac{4}{3}.\frac{3}{2}=\frac{101}{2}\)

Answer 2

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

\(=\frac{101}{100}.\frac{100}{99}.....\frac{4}{3}.\frac{3}{2}=\frac{101}{2}\)

Answer 3

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

  \(A=\frac{101}{100}\cdot\frac{100}{99}\cdot....\cdot\frac{4}{3}\cdot\frac{3}{2}\)

  \(A=\frac{101}{\frac{4}{2}}=\frac{101}{2}\)