Giúp mình với hiccc

a) = 29/15

b) = 7/15

c) = 1

d) = 3

e) = 67/17

f) = 2

mk nhanh nhất tk cho mk nha

a/\(\frac{3}{5}+\frac{4}{3}=\frac{9}{15}+\frac{20}{15}=\frac{29}{15}\)

b/\(\frac{2}{3}-\frac{1}{5}=\frac{10}{15}-\frac{3}{15}=\frac{7}{15}\)

c/\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{7}{6}\)

d,\(\frac{3}{5}+\frac{4}{7}+\frac{7}{5}+\frac{3}{7}=\left(\frac{3}{5}+\frac{7}{5}\right)+\left(\frac{4}{7}+\frac{3}{7}\right)=2+1=3\)

a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)

`!`

\(a,\dfrac{8}{9}\times\dfrac{7}{5}-\dfrac{7}{9}\times\dfrac{2}{5}+\dfrac{2}{9}\\ =\dfrac{8}{9}\times\left(\dfrac{7}{5}-\dfrac{2}{5}\right)+\dfrac{2}{9}\\ =\dfrac{8}{9}\times\dfrac{5}{5}+\dfrac{2}{9}\\ =\dfrac{8}{9}+\dfrac{2}{9}\\ =\dfrac{10}{9}\)

\(b,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}\\ =\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}\\ =\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\\ =\dfrac{1}{4}\times\dfrac{9}{9}\\ =\dfrac{1}{4}\)

ai giúp mik ik T_T

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

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

= - \(\dfrac{47}{33}\)

a=8/7

b=22/9

c=73/126

a, 1-3/7+4/7

 = 1-7/7

 = 1-1

 = 0

b,3-7/9+2/9

 = 3- 9/9

 = 3- 1

 = 2

c,=73/126

a) 3 2/7 - 2/5 + 5/7 - 3/5

= 3 + 2/7 - (2/5 + 3/5) + 5/7

= 3 + (2/7 + 5/7) - 1

= 3 + 1 - 1

= 3

b) 5 4/7 - 4/9 + 1 3/7 - 5/9

= 5 + 4/7 - 4/9 + 1 + 3/7 - 5/9

= (5 + 1) + (4/7 + 3/7) - (4/9 + 5/9)

= 6 + 1 - 1

= 6

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

\(=\left(\dfrac{23}{7}+\dfrac{5}{7}\right)-\left(\dfrac{2}{5}+\dfrac{3}{5}\right)\)

\(=4-1\)

\(=3\)

b) \(5\dfrac{4}{7}-\dfrac{4}{9}+1\dfrac{3}{7}-\dfrac{5}{9}\)

\(=\left(\dfrac{32}{7}+\dfrac{10}{7}\right)-\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)

\(=6-1\)

\(=5\)

a) x × 7/9 = 2/3 + 2 1/2

x × 7/9 = 2/3 + 5/2

x × 7/9 = 19/6

x = 19/6 : 7/9

x = 57/14

b) 5/7 + x : 9/4 = 4/3

x : 9/4 = 4/3 - 5/7

x : 9/4 = 13/21

x = 13/21 × 9/4

x = 39/28

Một tốp thợ có 120 người dự địnhnlamf trong 50 ngày . Khi bắt đầu làm có một số người đến làm thêm nên làm xong công việc đó trong 30 ngày . Hỏi có bao nhiêu người đến thêm?

Ta có: \(A=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{99\cdot101}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=\dfrac{1}{3}-\dfrac{1}{101}\)

\(=\dfrac{98}{303}\)

Với công thức \(\dfrac{a}{x.\left(x+a\right)}=\dfrac{a}{x}-\dfrac{a}{x+a}\)

Ta có: \(A=\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{99}-\dfrac{2}{101}\)

\(=\dfrac{2}{3}-\dfrac{2}{101}\)

xàm??

.-. ?

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