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7 tháng 9 2021

\(\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{32}{40}+\dfrac{48}{56}+\dfrac{14}{21}\\ =\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{4}{5}+\dfrac{6}{7}+\dfrac{2}{3}\\ =\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{7}+\dfrac{6}{7}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)\\ =1+1+1=3\)

AH
Akai Haruma
Giáo viên
7 tháng 9 2021

Lời giải:

$\frac{1}{3}+\frac{1}{7}+\frac{1}{5}+\frac{32}{40}+\frac{48}{56}+\frac{14}{21}$

$=\frac{1}{3}+\frac{1}{7}+\frac{1}{5}+\frac{4}{5}+\frac{6}{7}+\frac{2}{3}$

$=(\frac{1}{3}+\frac{2}{3})+(\frac{1}{7}+\frac{6}{7})+(\frac{1}{5}+\frac{4}{5})$

$=\frac{3}{3}+\frac{7}{7}+\frac{5}{5}=1+1+1=3$

6 tháng 11 2021

D

23 tháng 2 2023

\(1\dfrac{4}{5}+2\dfrac{5}{7}+3\dfrac{4}{5}+4\dfrac{5}{7}\)

\(\text{=}\left(1\dfrac{4}{5}+3\dfrac{4}{5}\right)+\left(2\dfrac{5}{7}+4\dfrac{5}{7}\right)\)

\(\text{=}1+3+\left(\dfrac{4}{5}+\dfrac{4}{5}\right)+2+4+\left(\dfrac{5}{7}+\dfrac{5}{7}\right)\)

\(\text{=}10+\dfrac{8}{5}+\dfrac{10}{7}\text{=}131\dfrac{1}{35}\)

5 tháng 8 2023

\(\dfrac{15}{14}\)\(\dfrac{10}{21}\) \(\times\) \(\dfrac{1}{5}\) = \(\dfrac{15}{14}\) \(\times\) \(\dfrac{21}{10}\) \(\times\) \(\dfrac{1}{5}\) = \(\dfrac{5\times3\times7\times3}{7\times2\times10\times5}\) = \(\dfrac{9}{20}\)

\(\times\) \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) = 1 + \(\dfrac{1}{5}\) = \(\dfrac{6}{5}\)

7 : \(\dfrac{1}{5}\) - \(\dfrac{1}{5}\) = 35 - \(\dfrac{1}{5}\) = \(\dfrac{174}{5}\)

6 + \(\dfrac{1}{5}\): 2 = 6 + \(\dfrac{1}{10}\) = \(\dfrac{61}{10}\) 

8 - \(\dfrac{1}{5}\) \(\times\) 7 = 8 - \(\dfrac{7}{5}\) = \(\dfrac{33}{5}\)

\(\dfrac{15}{14}\) : \(\dfrac{10}{21}\) x \(\dfrac{1}{5}\)   =   \(\dfrac{15}{14}\) x \(\dfrac{21}{10}\) x \(\dfrac{1}{5}\)  =   \(\dfrac{9}{4}\) x \(\dfrac{1}{5}\)  =  \(\dfrac{9}{20}\)

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

7 : \(\dfrac{1}{5}-\dfrac{1}{5}\)  =  \(\dfrac{7}{1}\) x \(\dfrac{5}{1}-\dfrac{1}{5}\)   =  \(\dfrac{35}{1}\) - \(\dfrac{1}{5}\)   =  \(\dfrac{175}{5}\) - \(\dfrac{1}{5}\)  =  \(\dfrac{174}{5}\)

6 + \(\dfrac{1}{5}\) : 2   =  \(\dfrac{6}{1}\) + \(\dfrac{1}{5}\) x \(\dfrac{1}{2}\)  =  \(\dfrac{6}{1}+\dfrac{1}{10}\)  =  \(\dfrac{60}{10}\) + \(\dfrac{1}{10}\)  = \(\dfrac{61}{10}\)

8 - \(\dfrac{1}{5}\) x 7  =  \(\dfrac{8}{1}\) - \(\dfrac{1}{5}\) x \(\dfrac{7}{1}\)  =  \(\dfrac{8}{1}-\dfrac{7}{5}\)  =  \(\dfrac{40}{5}\) - \(\dfrac{7}{5}\) = \(\dfrac{33}{5}\)

Sai Báo Lại Mình Nha!

8 tháng 1 2023

a)

\(=\dfrac{13}{5}+\dfrac{7}{5}\cdot\dfrac{7}{2}\)

\(=\dfrac{13}{5}+\dfrac{49}{10}\\ =\dfrac{26}{10}+\dfrac{49}{10}\\ =\dfrac{15}{2}\)

b)

\(=\dfrac{52}{4}-\dfrac{11}{3}:\dfrac{7}{6}\)

\(=\dfrac{52}{4}-\dfrac{22}{7}\\ =\dfrac{69}{7}\)

TC
Thầy Cao Đô
Giáo viên VIP
8 tháng 1 2023

a) $2\dfrac35 + 1\dfrac25 . 3\dfrac12$

$= \dfrac{13}5 + \dfrac75.\dfrac72$

$= \dfrac{26}{10} + \dfrac{49}{10}$

$=\dfrac{15}2$.

b) $4\dfrac34 - 3\dfrac23 : 1\dfrac16$

$= \dfrac{19}4 - \dfrac{11}3 : \dfrac76$

$= \dfrac{19}4 - \dfrac{11}3 . \dfrac67$

$= \dfrac{19}4 - \dfrac{22}7$

$= \dfrac{45}{28}$.

8 tháng 9 2023

\(\dfrac{13}{2}=6,5=\dfrac{65}{10}\)

\(\dfrac{11}{40}=0,275=\dfrac{275}{1000}\)

\(\dfrac{32}{5}=6,4=\dfrac{64}{10}\)

\(\dfrac{21}{250}=0,084=\dfrac{84}{1000}\)

\(\dfrac{1}{200}=0,005=\dfrac{5}{1000}\)

AH
Akai Haruma
Giáo viên
30 tháng 4 2023

Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$

$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$

$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$

AH
Akai Haruma
Giáo viên
30 tháng 4 2023

Bài 2:

$18: \frac{x\times 0,4+0,32}{x}+5=14$

$18: \frac{x\times 0,4+0,32}{x}=14-5=9$

$\frac{x\times 0,4+0,32}{x}=18:9=2$

$x\times 0,4+0,32=2\times x$

$2\times x-x\times 0,4=0,32$

$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$

LM
Lê Minh Vũ
CTVHS VIP
5 tháng 6 2023

\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)

\(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)

\(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)

\(\dfrac{40}{14}=\dfrac{20}{7}\)

\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)

=\(\dfrac{9}{2}+\dfrac{1}{11}\)

=\(\dfrac{101}{22}\)

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)

\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)

\(x=\dfrac{102}{21}=\dfrac{34}{7}\)