Question

Tính tích P=\(1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{99}\right)\)

Lưu ý:dấu chấm là dấu nhân

Answer 1

Ta có: \(1-\frac{1}{2}.\frac{2}{3}....\frac{98}{99}\)

 = \(1-\frac{1}{99}\)

=    \(\frac{98}{99}\)

Answer 2

P=\(1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).........\left(1-\frac{1}{99}\right)\)

P=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.........\frac{98}{99}=\frac{1}{99}\)

Answer 3

\(P=1-\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{99}=1-\frac{1}{99}=\frac{98}{99}\)

Nha ban Speed Ninja

Answer 4

P=1-\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{99}\right)\)

P=\(1-\frac{1}{2}.\frac{2}{3}...\frac{98}{99}\)

P=1-1/99=98/99

Answer 5

P= 1-1/2x2/3x...x98/99

P= 1/2x2/3x...x98/99

P= 1/99

Answer 6

1-1/2.2/3...98/99=1-98/99=1/99

Answer 7

\(P=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{99}\right)=1-\frac{1}{2}.\frac{2}{3}....\frac{98}{99}=1-\frac{1}{99}=\frac{98}{99}\)