A = 2/99 nhé !

kết  bạn với mình nha mình hết lượt  rồi

\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)....\left(1-\frac{1}{99}\right)\)

\(=\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{4}{4}-\frac{1}{4}\right)\left(\frac{5}{5}-\frac{1}{5}\right)....\left(\frac{99}{99}-\frac{1}{99}\right)\)

\(=\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{98}{99}=\frac{2}{99}\)

\(=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\)x \(\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\) x\(\left(\dfrac{4}{4}+\dfrac{1}{4}\right)\) x \(\left(\dfrac{5}{5}+\dfrac{1}{5}\right)\)

\(=\dfrac{3}{2}\) x \(\dfrac{4}{3}\) x \(\dfrac{5}{4}\) x \(\dfrac{6}{5}\) 

\(=3\)

= 2

3

A=9954

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Dấu chấm là nhân

a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)

\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)

CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)

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

\(=\frac{98}{99}\)

Tìm X

\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\)

\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\)

\(=\frac{1}{5}\)

\(\Rightarrow x-100=\frac{1}{5}\)

\(x=\frac{1}{5}+100\)

\(x=\frac{1}{5}+\frac{500}{5}\)

\(x=\frac{501}{5}\)

khó wá ko giải đc 

=1/2x2/3x/3/4/.................x2003/2004

= 1x2x3x................x2003

    ___________________

2x3x4x................x 2004

=1/2004

(1 - 1/2) . (1 - 1/3 ). (1- 1/4) . (1- 1/5).  ....  . (1- 1/2004)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2003}{2004}\)

\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot...\cdot2003\right)}{\left(2\cdot3\cdot4\cdot5\cdot....\cdot2003\right)\cdot2004}\)

\(=\frac{1}{2004}\)

a) (x-3)+(x-2)+(x-1)+....+10+11=11

(x-3)+(x-2)+(x-1)+....+10      =0

gọi số hạng của tổng vế trái là  n

(x-3+10).\(\frac{n}{2}\)=0

(x+7).n:2=0

(x+7)  =0

\(\Rightarrow\)x+7=0           (do n\(\ne\)0)

       x=0-7

       x=-7

b) \(\frac{2}{3}\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right]<=x<=4\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{6}\right]\)

     \(\frac{2}{3}.\frac{11}{12}<=x<=\frac{13}{3}.\frac{1}{3}\)

      \(\frac{11}{18}<=x<=\frac{13}{9}\)

do x\(\in\)z nên x=1

vậy x=1