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

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

\(=\frac{1}{99}\)

cám ơn bn:)

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

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

\(P=\frac{1.2.3...98}{2.3.4...99}=\frac{1}{99}\)

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

\(\Rightarrow P=\frac{1}{2}.\frac{2}{3}.........\frac{98}{99}\)\(=\frac{1.2.3...98}{2.3.4...99}=\frac{1}{99}\)

Vậy \(P=\frac{1}{99}\)

 1/1*2+1/2*3+1/3*4+.....................+1/98*99+1/99*100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

đầu tiên ta thấy:

1/1*2=1/2=1-1/2 từ đó ta tách ra sau đó dùng pp khử 

Giúp mình vs

P=(1-1/2)(1-1/3)(1-1/4)....(1-1/100)

P=1/2.2/3.3/4.......99/100

P=(1.2.3....99)/(2.3.4...100)=1/100

Vậy P=1/100

p=1/100

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

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{99}.\frac{99}{100}\)

\(\Leftrightarrow P=\frac{1}{100}\)

P=1/2 * 2/3 *3/4 *....* 98/99

P=(1.2.3.4....98)/(2.3.4....99)

=> P=1/99

Đáp số: P=1/99

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

P = \(\frac{1}{2}\). \(\frac{2}{3}\). \(\frac{3}{4}\).  . . . . \(\frac{99}{100}\)

P = ( 1.2.3. ... . 99) : (2.3.4. ... .100)

P = 1 : 100

P = \(\frac{1}{100}\)

 Vậy ................

~~ học tốt ~~

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

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

= \(\frac{1}{99}\)
 

Tại sao lại bằng 1/ 99 luôn? 

P =  1 − 2 1 1 − 3 1 ... 1 − 99 1 =  2 1 . 3 2 ..... 99 98 =  99 1