Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(1-1/2) x (1-1/3) x .... x (1-1/100)
= 1/2 x 2/3 x ... x 99/100
= 1x2x...x99/2x3x..x100
= 1/ 100
a, \(\frac{3}{5}+25-\frac{1}{5}\)
\(=\left(\frac{3}{5}-\frac{1}{5}\right)+25\)
\(=\frac{2}{5}+25\)
\(=25\frac{2}{5}\)
\(=25,4\)
b, \(13.3.32,27+67,63.39\)
\(=39.32,27+67,63.39\)
\(=39\left(32,37+67,63\right)\)
\(=39.100\)
\(=3900\)
c, \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{99}{100}\)
\(=\frac{1}{100}\)
Bài 3 :suy ra thì 2 số cùng thêm 1,5 nên hiệu k thay đổi
Số lớn : 3 lần
Số bé : 1 lần
hiệu số phần bằng nhau là :
3 - 1 = 2 ( phần )
số lớn LS là :
22,8 : 2 x 3 34,2
số lớn LĐ là :
34,2 - 1,5 =32,7
số bé là :
32,7 - 22,8 = 9,9
đúng 100%
= 1/1x2 + 1/2x3 + 1/3x4 ...... +1/9x10
= 1-1/2+1/2-1/3+1/3-1/4+........+1/9-1/10
=1-1/10=9/10
đặt A=1/1 x 1/2 + 1/2 x 1/3 + 1/3 + 1/4 + .......... + 1/9 x 1/10
\(A=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+...+\frac{1}{9}\cdot\frac{1}{10}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
đặt B=2/1 x 2 + 2/2 x 3 + 2/3 x4 + .............. + 2/98 x 99 + 2/99 x 100
\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)\)
\(=2\times\frac{99}{100}\)
\(=\frac{99}{50}\)
a) A = 2 + 4 + 6 + 8 + ... + 1000
Ta có : A = 2 + 4 + 6 + 8 + ... + 1000 ( có 500 số )
= (1000 + 2) . 500 : 2 = 250500
c) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
Bài 1:
a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)
\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{3}\)
\(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)
\(\dfrac{1}{4}\) \(x\)= \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)
\(\dfrac{1}{4}x=\dfrac{11}{24}\)
\(x=\dfrac{11}{24}:\dfrac{1}{4}\)
\(x=\dfrac{11}{24}\times4\)
\(x=\dfrac{11}{6}\)
b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)
\(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)
\(x\) = \(\dfrac{9}{10}\)
\(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\times...\times1\frac{1}{100}\)
\(=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times...\times\frac{101}{100}\)
\(=\frac{3\times4\times5\times...\times101}{2\times3\times4\times...\times100}\)
\(=\frac{101}{2}\)
a, = 1/2 x 2/3 x 3/4 x .... x 99/100 = 1/100
b, = 24/25 x 5/7 x 7/9 x .... x 97/99 = 24/25 x 5/99 = 8/165
a) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
=