a) =

b) =

c) =

d) = .

a) \(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{6}{14}+\dfrac{7}{17}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{13^2}{12^2}=\dfrac{169}{144}\)

b)\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2=\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2=\left(\dfrac{-1}{12}\right)^2=\dfrac{\left(-1\right)^2}{12^2}=\dfrac{1}{144}\)

c)\(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.5^4.2^8}{5^{10}.2^{10}}=\dfrac{5^8.2^8}{5^8.5^2.2^8.2^2}=\dfrac{1}{5^2.2^2}=\dfrac{1}{25.4}=\dfrac{1}{100}\)

d)\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4=\dfrac{\left(-10\right)^5.\left(-6\right)^4}{3^5.5^4}=\dfrac{\left(-2\right)^5.5^5.2^4.3^4}{3^4.3.5^4}=\dfrac{\left(-2\right)^5.5.5^42^4}{3.5^4}=\dfrac{\left(-2\right)^5.5.2^4}{3}=\dfrac{-2560}{3}=-853\dfrac{1}{3}\)

a.\(12,5.\left(-\dfrac{5}{7}\right)+1,5.\left(-\dfrac{5}{7}\right)\)

\(=\left(-\dfrac{5}{7}\right).\left(12,5+1,5\right)\)

\(=-10\)

b,\(\left(-\dfrac{2}{5}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)

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

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

\(=-\dfrac{3}{4}\)

c,\(12.\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)

\(=12.\dfrac{4}{9}+\dfrac{4}{3}\)

\(=\dfrac{16}{3}+\dfrac{4}{3}\)

\(=\dfrac{20}{3}\)

d,\(1:\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{1}{1}:\dfrac{1}{144}\)

\(=144\)

e,\(15.\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15.\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

a) = ( 12,5 +1,5 ). \(\left(-\dfrac{5}{7}\right)\)

= 14 . \(\left(-\dfrac{5}{7}\right)\)

= -10

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

= \(\left(-\dfrac{3}{5}+0\right)\): \(\dfrac{4}{5}\)

= \(\dfrac{3}{4}\)

c) = \(\left(12.-\dfrac{2}{9}\right)\) + \(\dfrac{4}{3}\)

= \(\dfrac{8}{3}\) + \(\dfrac{4}{3}\)

= \(-\dfrac{4}{3}\)

d) = 1: \(\dfrac{23}{48}\)

=\(\dfrac{48}{23}\)

e) =\(\left(15.-\dfrac{2}{9}\right)-\dfrac{7}{3}\)

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

=\(-\dfrac{17}{3}\)

f) = 10 485.76

a, \(12,5.\left(\dfrac{-5}{7}\right)+1,5.\left(\dfrac{-5}{7}\right)\)

\(=\left(\dfrac{-5}{7}\right).\left(12,5+1,5\right)=\left(\dfrac{-5}{7}\right).14=-10\)

b, \(\left(\dfrac{-2}{5}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)

\(=\left[\left(-\dfrac{2}{5}-\dfrac{1}{5}\right)+\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)\right]:\dfrac{4}{5}\)

\(=\left(\dfrac{-3}{5}+0\right):\dfrac{4}{5}=\dfrac{-3}{5}.\dfrac{5}{4}=\dfrac{-3}{4}\)

c, \(12.\left(\dfrac{-2}{3}\right)^2+\dfrac{4}{3}\)

\(=12.\left(\dfrac{-2^2}{3^2}\right)+\dfrac{4}{3}=12.\left(\dfrac{4}{9}\right)+\dfrac{4}{3}=\dfrac{16}{3}+\dfrac{4}{3}=5\)

d, \(1:\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2=1:\left(\dfrac{-1}{12}\right)^2=1:\dfrac{1}{144}=144\)

e, \(15.\left(\dfrac{-2}{3}\right)^2-\dfrac{7}{3}\)

\(=15.\dfrac{4}{9}-\dfrac{7}{3}=\dfrac{20}{3}-\dfrac{7}{3}=\dfrac{13}{3}\)

a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\left(\dfrac{4}{1}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\left(\dfrac{20}{5}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\dfrac{8}{5}.\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=1-\dfrac{2}{5}.\dfrac{25}{-4}\)

\(=1-\dfrac{-5}{2}\)

\(=\dfrac{2}{2}-\dfrac{-5}{2}\)

\(=\dfrac{7}{2}\)

dài quá nên mik sẽ giải lần lượt mỗi câu trả lời là một câu nhá bạn!!

Giải:

a)(4-12/5).25/8-2/5:-4/25

=8/5.25/8-(-5/2)

=5+5/2

=15/2

b)(-5/24+3/4-7/12):(-5/16)

=-1/24:(-5/16)

=2/15

c)6/7+5/4:(-5)-(-1/28).(-2)²

=6/7+(-1/4)-(-1/28).4

=6/7-1/4-(-1/7)

=6/7-1/4+1/7

=(6/7+1/7)-1/4

=1-1/4

=3/4

Chúc bạn học tốt!

Toàn câu dễ nên bạn tự làm đi.

Trong lúc bạn đánh xong bài này thì bạn có thể làm xong rồi đó.

Đừng có ỷ lại vào người khác ,động não lên.

a: \(=\dfrac{5^8\cdot2^8}{5^{10}\cdot2^{10}}=\dfrac{1}{100}\)

b: =7/2+2/7

=49/14+4/14

=53/14

c: \(=\dfrac{11}{12}\cdot\dfrac{16}{33}\cdot\dfrac{3}{5}=\dfrac{1}{3}\cdot\dfrac{3}{5}\cdot\dfrac{4}{3}=\dfrac{1}{5}\cdot\dfrac{4}{3}=\dfrac{4}{15}\)

d: \(=15\cdot\dfrac{4}{9}\cdot\dfrac{-7}{3}=\dfrac{15}{27}\cdot\left(-28\right)=-28\cdot\dfrac{5}{9}=-\dfrac{140}{9}\)

g: \(=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{9}{45}+\dfrac{36}{45}\right)=1+1=2\)

j: =12*4/9+4/3

=16/3+4/3

=20/3

a) \(\left(1,25\right)^3.8^3=\left(1,25.8\right)^3=1000\)

b) \(\left(\dfrac{-11}{9}\right)^4.\left(\dfrac{27}{22}\right)^4=\left(\dfrac{-11}{9}.\dfrac{27}{22}\right)^4=\left(\dfrac{-11.9.3}{9.2.\left(-11\right)}\right)^4\)

\(=\left(\dfrac{3}{2}\right)^4=\dfrac{81}{16}\)

c) \(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)

d) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=100^{-1}=0,01\)

a: \(=\left(\dfrac{19}{6}-\dfrac{2}{5}\right):\left(\dfrac{29}{6}+\dfrac{7}{10}\right)\)

\(=\dfrac{19\cdot5-2\cdot6}{30}:\dfrac{290+42}{30}=\dfrac{83}{332}=\dfrac{1}{4}\)

b: \(=\dfrac{\left(\dfrac{102}{25}-\dfrac{2}{25}\right)\cdot\dfrac{17}{4}}{\left(6+\dfrac{5}{9}-3-\dfrac{1}{4}\right)\cdot\dfrac{16}{7}}\)

\(=\dfrac{4\cdot\dfrac{17}{4}}{\dfrac{16}{7}\cdot\dfrac{119}{36}}=\dfrac{17}{\dfrac{68}{9}}=17\cdot\dfrac{9}{68}=\dfrac{9}{4}\)

c: \(=\left(\dfrac{120}{60}-\dfrac{15}{60}+\dfrac{20}{60}-\dfrac{36}{60}\right):\left(\dfrac{45}{15}-\dfrac{3}{15}-\dfrac{25}{15}\right)\)

\(=\dfrac{89}{60}:\dfrac{17}{15}=\dfrac{89}{60}\cdot\dfrac{15}{17}=\dfrac{89}{68}\)

\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)

\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)

\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)

a) `(5^4 . 20^4)/(25^3 .4^5)`

`=(5^4 . (5.4)^4)/((5^2)^3 .4^5)`

`= (5^4 . 5^4 . 4^4)/(5^6 . 4^5)`

`= (5^2)/4=25/4`

b) `(-10/3)^5 . (-6/5)^4`

`=-10/3 . [(-10/3) . (-6/5)]^4`

`= -10/3 . [ (-5.2 . (-2).3)/(3.5)]^4`

`=-10/3 . 4^4`

`=-2560/3`

A) \(=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=\dfrac{1}{100}\)

B)=\(\left(\dfrac{-10}{3}\right).\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4\)

   =\(\left(\dfrac{-10}{3}\right).\left(\dfrac{-10}{3}.\dfrac{-6}{5}\right)^4\)

   =\(\left(\dfrac{-10}{3}\right).\left(\dfrac{60}{15}\right)^4\)

   =\(\left(\dfrac{-10}{3}\right).4^4\)

   =\(\left(\dfrac{-10}{3}\right).256\)

   =\(\dfrac{-2650}{3}\)

a) \(\dfrac{5^4.20^4}{25^3.4^5}=\dfrac{5^4.4^4.5^4}{5^6.4^5}=\dfrac{5^2}{4}=\dfrac{25}{4}\) 

b) \(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-10}{3}\right).\left(\dfrac{-6}{5}\right)^4=\left(\dfrac{-10}{3}.\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)=\left(4\right)^4.\left(\dfrac{-10}{3}\right)=256.\left(\dfrac{-10}{3}\right)=\dfrac{-2560}{3}\)

a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)

b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)

c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)

d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)