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NV
3 tháng 12 2022

\(A=\dfrac{7}{3.6}+\dfrac{7}{6.9}+...+\dfrac{7}{96.99}\)

\(=\dfrac{7}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{96.99}\right)\)

\(=\dfrac{7}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{96}-\dfrac{1}{99}\right)\)

\(=\dfrac{7}{3}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=\dfrac{224}{297}\)

26 tháng 2 2023

A = \(\dfrac{7}{3\times6}\) + \(\dfrac{7}{6\times9}\) + \(\dfrac{7}{9\times12}\) + \(\dfrac{7}{12\times15}\)+ .....+\(\dfrac{7}{96\times99}\)

A = \(\dfrac{7}{3}\) x ( \(\dfrac{3}{3\times6}\) + \(\dfrac{3}{6\times9}\)\(\dfrac{3}{9\times12}\)\(\dfrac{3}{12\times15}\)+......+\(\dfrac{3}{96\times99}\))

A = \(\dfrac{7}{3}\) x ( \(\dfrac{1}{3}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{12}\)\(\dfrac{1}{12}\) - \(\dfrac{1}{15}\)+....+ \(\dfrac{1}{96}\) - \(\dfrac{1}{99}\))

A = \(\dfrac{7}{3}\) x ( \(\dfrac{1}{3}\)\(\dfrac{1}{99}\))

A = \(\dfrac{224}{297}\)

1 tháng 7 2021

\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)

1 tháng 7 2021

1. Tính 

\(a,5\times\dfrac{7}{3}=\dfrac{35}{3}\)

\(b,\dfrac{13}{4}:7=\dfrac{13}{4}\times\dfrac{1}{7}=\dfrac{13}{28}\)

2. Tính

\(a,\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}\)

\(=\dfrac{15}{35}+\dfrac{14}{35}+\dfrac{3}{4}\)

\(=\dfrac{29}{35}+\dfrac{3}{4}\)

\(=\dfrac{116}{140}+\dfrac{105}{140}\)

\(=\dfrac{221}{140}\)

\(b,\dfrac{9}{7}-\dfrac{5}{11}\times\dfrac{11}{7}\)

\(=\dfrac{9}{7}-\dfrac{55}{77}\)

\(=\dfrac{99}{77}-\dfrac{55}{77}\)

\(=\dfrac{44}{77}=\dfrac{4}{7}\)

\(c,\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{4}{7}\)

\(=\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}\right)\)

\(=\dfrac{3}{5}\times\dfrac{9}{7}\)

\(=\dfrac{27}{35}\)

\(d,\dfrac{7}{9}\times\dfrac{2}{5}:\dfrac{3}{11}\)

\(=\dfrac{14}{45}:\dfrac{3}{11}\)

\(=\dfrac{14}{45}\times\dfrac{11}{3}\)

\(=\dfrac{154}{135}\)

\(e,\dfrac{9}{7}+\dfrac{2}{3}-\dfrac{1}{4}\)

\(=\dfrac{27}{21}+\dfrac{14}{21}-\dfrac{1}{4}\)

\(=\dfrac{41}{21}-\dfrac{1}{4}\)

\(=\dfrac{164}{84}-\dfrac{21}{84}\)

\(=\dfrac{143}{84}\)

\(g,\dfrac{4}{9}:\dfrac{3}{5}\times\dfrac{2}{11}\)

\(=\dfrac{4}{9}\times\dfrac{5}{3}\times\dfrac{2}{11}\)

\(=\dfrac{20}{27}\times\dfrac{2}{11}\)

\(=\dfrac{40}{297}\)

\(h,\dfrac{7}{2}-\dfrac{3}{10}:\dfrac{2}{5}\)

\(=\left(\dfrac{7}{2}-\dfrac{3}{10}\right):\dfrac{2}{5}\)

\(=\left(\dfrac{35}{10}-\dfrac{3}{10}\right):\dfrac{2}{5}\)

\(=\dfrac{32}{10}:\dfrac{2}{5}\)

\(=\dfrac{16}{5}\times\dfrac{5}{2}\)

\(=\dfrac{80}{10}=8\)

6 tháng 11 2021

D

16 tháng 10 2023

`(6/11 +5/11) xx 3/7`

`= 11/11xx 3/7`

`=1xx3/7`

`=3/7`

__

`3/5 xx 7/9 - 3/5 xx 2/9`

`= 3/5 xx (7/9-2/9)`

`= 3/5 xx 5/9`

`= 15/45`

`= 1/3`

__

`(6/7 -4/7):2/5`

`= 2/7 : 2/5`

`= 2/7 xx 5/2`

`= 10/14`

`= 5/7`

16 tháng 10 2023

\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}\)

\(=\dfrac{11}{11}\times\dfrac{3}{7}\\ =1\times\dfrac{3}{7}=\dfrac{3}{7}\)

_____

\(\dfrac{3}{5}\times\dfrac{7}{9}-\dfrac{3}{5}\times\dfrac{2}{9}\\ =\dfrac{3}{5}\times\left(\dfrac{7}{9}-\dfrac{2}{9}\right)\\ =\dfrac{3}{5}\times\dfrac{5}{9}\\ =\dfrac{3}{9}\\ =\dfrac{1}{3}\)

_________

\(\left(\dfrac{6}{7}-\dfrac{4}{7}\right):\dfrac{2}{5}\\ =\dfrac{2}{7}\times\dfrac{5}{2}\\ =\dfrac{10}{14}\\ =\dfrac{5}{7}\)

8 tháng 6 2023

Bài 3

a,26/100+0,009+41/100+0,24

0,26+0,09+0,41+0,24

(0,26+0,24)+(0,09+0,41)

0,5+0,5

=1

b,9+1/4+6+2/7+7+3/5+8+2/3+2/5+1/3+5/7+3/4

(9+6+7+8)+(2/7+5/7)+(1/4+3/4)+(3/5+2/5)+(2/3+1/3)

30+1+1+1+1

=34

Bài 4,5 khó quá mik ko bít lamf^^))

 

 

16 tháng 9 2023

Bài 4: a, \(\dfrac{2008}{2009}\) < 1; \(\dfrac{10}{9}\) > 1

           \(\dfrac{2008}{2009}\) < \(\dfrac{10}{9}\)

         b, \(\dfrac{1}{a+1}\) và \(\dfrac{1}{a-1}\)

Ta có: a + 1 > a - 1 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)

 

24 tháng 6 2021

`a)1/7xx2/7+1/7xx5/7+6/7`

`=1/7xx(2/7+5/7)+6/7`

`=1/7xx1+6/7`

`=1/7+6/7=1`

`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`

`=6/11xx(4/9+7/9-2/9)`

`=6/11xx9/9`

`=6/11`

24 tháng 6 2021

Sorry nãy ghi thiếu.

`c)4/25xx5/8xx25/4xx24`

`=(4xx5xx25xx24)/(25xx8xx4)`

`=(4xx5xx24)/(4xx8)`

`=(5xx24)/8`

`=5xx3=15`

28 tháng 3 2022

ai giúp mik ik T_T

5 tháng 8 2023

a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)

\(\dfrac{1}{13}\) \(\times\)\(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))

\(\dfrac{1}{13}\) \(\times\)\(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))

\(\dfrac{1}{13}\) \(\times\)  \(\dfrac{317}{73}\)

\(\dfrac{317}{949}\)

b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)

=   \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)

\(\dfrac{199}{45}\)

c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

=   \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

\(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)

\(\dfrac{20696}{315}\)

d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)

= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)

= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))

= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)

= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)

e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))

= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)

=  -1 - \(\dfrac{2}{11}\)  - \(\dfrac{8}{33}\)

\(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) -  \(\dfrac{8}{33}\)

= - \(\dfrac{47}{33}\)

27 tháng 3 2022

43/14

1/3

71/15

27 tháng 3 2022

\(\dfrac{43}{14}\)

\(\dfrac{1}{3}\)

\(\dfrac{71}{15}\)

3 tháng 8 2016

gcjjjjjjjjjjjhm.f

17 tháng 11 2021

Vũ Thị Trang lại là một nạn nhân tiếp tục bị báo cáo

8 tháng 9 2023

\(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+...+\dfrac{7}{90}\)

\(A=7x\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}\right)\)

\(A=7x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{9x10}\right)\)

\(A=7x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(A=7x\left(1-\dfrac{1}{10}\right)\)

\(A=7x\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)

\(A=7x\dfrac{9}{10}=\dfrac{63}{10}\)