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

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

c) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=\dfrac{1}{100}\)

d) \(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4=\dfrac{-10^5}{3^5}.\dfrac{-6^4}{5^4}=\dfrac{-\left(2.5\right)^5.\left(3.2\right)^4}{3^5.5^4}=\dfrac{-29.5}{3}=-853\dfrac{1}{3}\)

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

`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`

`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`

`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`

`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`

`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`

e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)

f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)

g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)

a: \(=\dfrac{5^4\cdot5^4\cdot2^8}{5^{10}\cdot2^{10}}=\dfrac{1}{5^2}\cdot\dfrac{1}{2^2}=\dfrac{1}{100}\)

b: \(=\left(\dfrac{12+8-3}{12}\right)\cdot\left(\dfrac{16-15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

Ta lần lượt có:

\(\dfrac{1}{0,25}=\dfrac{100}{25}=4;\) \(\left(1\dfrac{1}{4}\right)^2=\left(\dfrac{5}{4}\right)^2=\dfrac{25}{16};\)

\(\dfrac{1}{\left(\dfrac{4}{3}\right)^2}=\dfrac{1}{\dfrac{16}{9}}=\dfrac{9}{16};\) \(\left(\dfrac{5}{4}\right)^3=\dfrac{125}{64}\) và \(\dfrac{1}{\left(-\dfrac{2}{3}\right)^3}=\dfrac{1}{-\dfrac{8}{27}}=-\dfrac{27}{8}\)

Vậy \(P=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\left(-\dfrac{27}{8}\right)\)

\(=\dfrac{25}{4}-25.\dfrac{9}{16}.\dfrac{64}{125}.\dfrac{8}{27}\)

\(=\dfrac{25}{4}-\dfrac{32}{15}\)

\(=\dfrac{375-128}{60}=\dfrac{247}{60}\)

P=4.(25/16)+25.(9/16:25/16):(-27/8)

=25/4 +25.(9/25):(-27/8)

=25/4 +9:(-27/8)

=25/4 +(-8/3)

=43/12

a)2401/81

b)1024

c)125/1

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