a: \(=-\dfrac{5}{12}\cdot\dfrac{9}{20}\cdot\dfrac{7}{17}=-\dfrac{21}{272}\)

b: \(=\dfrac{13}{17}\left(-\dfrac{4}{5}-\dfrac{3}{4}\right)=\dfrac{-13}{17}\cdot\dfrac{31}{20}=-\dfrac{403}{340}\)

c: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{5}{7}=-\dfrac{5}{7}+\dfrac{5}{7}=0\)

d: \(=\dfrac{12}{5}-\dfrac{3}{10}=\dfrac{21}{10}\)

mn giúp mình đi

a: \(=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}=4-\dfrac{4}{7}=\dfrac{24}{7}\)

b: \(=\dfrac{11}{2}\cdot\dfrac{15}{4}=\dfrac{165}{8}\)

c: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}-6-\dfrac{2}{9}=6+\dfrac{3}{5}=\dfrac{33}{5}\)

d: \(=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}=5+\dfrac{7}{11}=\dfrac{62}{11}\)

Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha

1, 7A = 7+7^2+7^3+....+7^2008

6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1

=> A = (7^2008-1)/6

Tk mk nha

\(A=1+7+7^2+7^3+...+7^{2007}\)

\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)

\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)

\(\Rightarrow6A=7^{2008}-1\)

\(\Rightarrow A=\frac{7^{2008}-1}{6}\)

a: =91/105+60/105-101/105

=50/105=10/21

c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)

d: =2-2/9

=18/9-2/9

=16/9

e: =24/36-9/36+8/36

=23/36

g: =5/2+1/2

=3

thanks

Bài 1: Tìm \(x\) 

a; \(x-2\) + 7 = 1.3.(-9)

   \(x\) - 2 + 7 = 3.(-9)

   \(x\) - 2 + 7 = - 27

  \(x\)             = - 27 - 7 + 2

   \(x\)            = - 34 + 2

   \(x\)            = - 32

Vậy \(x=-32\)

Bài 1

c; - 2\(x\) + 5 = 7

     - 2\(x\)       = 7  - 5

    - 2\(x\)        = - 2

        \(x\)        = -2 : (-2)

        \(x\)        = - 1

Vậy \(x\)         = - 1

31/8.   3/4

20/9.     3/16

a: \(\dfrac{7}{8}+3=\dfrac{3\cdot8+7}{8}=\dfrac{31}{8}\)

b: \(2-\dfrac{5}{4}=\dfrac{2\cdot4-5}{4}=\dfrac{3}{4}\)

\(4\cdot\dfrac{5}{9}=\dfrac{20}{9}\)

d: \(\dfrac{3}{8}:2=\dfrac{3}{16}\)

a: \(B=\frac12-\left\lbrack\frac38+\left(-\frac74\right)\right\rbrack\)

\(=\frac12-\frac38+\frac74\)

\(=\frac48-\frac38+\frac{14}{8}=\frac{15}{8}\)

b: \(-\frac{4}{12}-\left(-\frac{13}{39}-0,25\right)+0,75\)

\(=-\frac13+\frac13+0,25+0,75\)

=0,25+0,75

=1

c: \(\frac12-\frac13+\frac{1}{23}+\frac16\)

\(=\left(\frac12-\frac13+\frac16\right)+\frac{1}{23}\)

\(=\left(\frac16+\frac16\right)+\frac{1}{23}=\frac13+\frac{1}{23}=\frac{26}{69}\)

d: \(\left(-\frac{13}{7}-\frac49\right)-\left(-\frac{10}{7}-\frac49\right)\)

\(=-\frac{13}{7}-\frac49+\frac{10}{7}+\frac49\)

\(=-\frac{13}{7}+\frac{10}{7}=-\frac37\)

e: \(\left(\frac78-\frac52+\frac47\right)-\left(-\frac37+1-\frac{13}{8}\right)\)

\(=\frac78-\frac52+\frac47+\frac37-1+\frac{13}{8}\)

\(=\frac{20}{8}-\frac52=0\)

f: \(-\frac37+\left(3-\frac34\right)-\left(2,25-\frac{10}{7}\right)\)

\(=-\frac37+3-\frac34-2,25+\frac{10}{7}\)

\(=\frac77+0,75-\frac34\)

=1
g: \(\left(\frac53-\frac37+9\right)-\left(2+\frac57-\frac23\right)+\left(\frac87-\frac43-10\right)\)

\(=\frac53-\frac37+9-2-\frac57+\frac23+\frac87-\frac43-10\)

\(=\left(\frac53+\frac23-\frac43\right)+\left(-\frac37-\frac57+\frac87\right)+\left(9-2-10\right)\)

\(=\frac33+7-10=1-3=-2\)