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\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot\dfrac{25}{24}\cdot\dfrac{36}{35}=\dfrac{12}{7}\)

9 tháng 12 2022

= 4/3*9/8*16/15*25/24*36/35

=2*2/1*3 * 3*3/2*4 *4*4/3*5 *5*5/4*6 * 6*6/5*7

= (2*3*4*5*6 / 1*2*3*4*5) * ( 2*3*4*5*6 / 3*4*5*6*7)

=6/1* 2/7 

= 12/7

 

1 tháng 11 2023

( 1 + 1/2 ) . ( 1+ 1/3) . ( 1+ 1/4 ) . ( 1+ 1/5 ) 

=3/2 . 4/3.5/4.6/5

= 3.4.5.6/2.3.4.5

=6/2 = 3 

1 tháng 11 2023

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

= 1 x \(\dfrac{77}{60}\)

\(\dfrac{77}{60}\)

Giải:

\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\) 

\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\) 

\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\) 

\(=\dfrac{1}{100}\)

20 tháng 11 2021

a)=2

b)=1/3

16 tháng 4 2022

\(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

\(=\left(\dfrac{1}{10}+\dfrac{9}{10}\right)+\left(\dfrac{2}{10}+\dfrac{8}{10}\right)+\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{4}{10}+\dfrac{6}{10}\right)+\dfrac{5}{10}\)

\(=1+1+1+1+\dfrac{5}{10}\)

\(=4+\dfrac{5}{10}\)

\(=\dfrac{45}{10}\)

\(13,25:0,5+13,25:0,25+13,25:0,125+13,25\times6\)

\(=13,25:\dfrac{1}{2}+13,25:\dfrac{1}{4}+13,25:\dfrac{1}{8}+13,25\times6\)

\(=13,25\times2+13,25\times4+13,25\times8+13,25\times6\)

\(=13,25\times\left(2+4+8+6\right)\)

\(=13,25\times20\)

\(=265\)

2 tháng 1 2023

\(a,=69,78\times\left(\dfrac{3}{4}+\dfrac{1}{4}+99\right)\\ =69,78\times\left(\dfrac{4}{4}+99\right)\\ =69,78\times\left(1+99\right)=69,78\times100=6978\\ b,34,1\times8+34,1\times0,5\\ =34,1\times\left(8+0,5\right)\\ =34,1\times8,5\\ =289,85\)

a: =69,78(3/4+1/4+99)

=69,78*100

=6978

b: =34,1(8+0,5)=34,1*8,5=289,85

23 tháng 5 2022

\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)

\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)

\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)

\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)

\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)

23 tháng 5 2022

(6 - 4\(\dfrac{1}{2}\) ) : 0,03 x 40%

= (6 - \(\dfrac{9}{2}\)) : \(\dfrac{3}{100}\) x \(\dfrac{40}{100}\)

\(\dfrac{3}{2}\) x\(\dfrac{100}{3}\) x \(\dfrac{40}{100}\)

= 20

23 tháng 5 2022

Bài khó vậy 

11 tháng 11 2021

11,6

 

11 tháng 11 2021

:v