bài 1b)

\(8\frac{1}{14}-6\frac37\)

C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)

C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)

\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)

bài 1 c)\(7-3\frac67\)

C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)

C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)

a) 1/2 + 3/4 - (3/4 - 4 - 5)

= 1/2 + 3/4 - 3/4 + 4 + 5

= (3/4 - 3/4) + (4 + 5) + 1/2

= 0 + 9 + 1/2

= 19/2

b) [9/16 + 8/(-27)] - (19/27- 7/16 - 2)

= 9/16 - 8/27 - 19/27 + 7/16 + 2

= (9/16 + 7/16) + (-8/27 - 19/27) + 2

= 1 - 1 + 2

= 2

c) -5/8 . [4/9 + 7/(-12)]

= -5/8 . (-5/36)

= 25/288

d) 7/10 . (-3/5) + 7/10 . (-2/5) - (-3/10)

= 7/10 . (-3/5 - 2/5) + 3/10

= 7/10 . (-1) + 3/10

= -2/5

e) -3/7 . 5/9 + 4/9 . (-3/7) + 2 3/7

= -3/7 . (5/9 + 4/9) + 17/7

= -3/7 . 1 + 17/7

= 2

f) 8 2/7 - (3 4/9 + 4 2/7)

= 8 + 2/7 - 3 - 4/9 - 4 - 2/7

= (8 - 3 - 4) + (2/7 - 2/7) - 4/9

= 1 - 4/9

= 5/9

h) 3.(-1/2)² - (4/5 + 8/15) : 5/6

= 3.1/4 - 4/3 : 5/6

= 3/4 - 8/5

= -17/20

DAI QUA . 

B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)

=\(11\frac34-6\frac56+4\frac12+1\frac23\)

=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)

=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)

=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)

=10+\(\frac{13}{12}\)

=\(\frac{120}{12}+\frac{13}{12}\)

=\(\frac{133}{12}\)

b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)

= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)

=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)

=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)

=\(\frac{108}{60}\)

=\(\frac95\)

a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)

   =1

b)=(1.2.3....8).(9-1-8)

   =(1.2.3....8).0

   =0

mik chỉ giải được zậy thôi.

t mik nha.

Hicc mình hông còn tgian mấy bạn giúp minh 2 câu thui cũng đc ạk

a)  5/9 + 4/9 . 3/7 + 4/9 . 4/7

  = 5/9 + 4/9 . (3/7 + 4/7)

  = 5/9 + 4/9 . 1

  = 5/9 + 4/9

  = 1 

Bài 1a

- \(\frac45\) : \(\frac87\)

= - \(\frac45\) x \(\frac78\)

= - \(\frac{7}{10}\)

Bài 1b:

- \(\frac38\). \(\frac79\) + \(\frac{-3}{8}\).\(\frac29\) + 0,25

= - \(\frac38\).(\(\frac79+\frac29\)) + 0,25

= - \(\frac38\). 1 + 0,25

= - 0,375 + 0,25

= - 0,125

Bài 1:

a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)

\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)

\(=\frac{-3}{5}\)

b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)

\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)

\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)

c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)

\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)

\(=3+2=5\)

d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)

\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)

\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)

e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)

\(=-1+1+\frac{-7}{9}\)

\(=-\frac{7}{9}\)

f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)

\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)

\(=1+\frac{-1}{7}=\frac{7}{7}+\frac{-1}{7}=\frac{6}{7}\)

\(\frac37+\frac{5}{13}+\frac{4}{13}\)

=\(\frac37+\left(\frac{5}{13}+\frac{4}{13}\right)\)

=\(\frac37+\frac{9}{13}\)

=\(\frac{39}{91}+\frac{63}{91}\)

=\(\frac{102}{91}\)

\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)

=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{7}{24}\right)\times\frac65+\frac12\)

=\(\frac{5}{24}\times\frac65+\frac12\)

=\(\frac14+\frac12\)

=\(\frac14+\frac24\)

=\(\frac34\)

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

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

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

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

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

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