\(A=\dfrac{6^{12}.3^3.5-7.9^7.2^{13}}{2.4^7.5-2^{14}.3^2}\)

\(=\dfrac{2^{12}.3^{12}.3^3.5-7.3^{14}.2^{12}}{2.2^{14}.5-2^{14}.3^2}\)

\(=\dfrac{2^{12}.3^{15}.5-7.3^{14}.2^{13}}{2.3^{14}.5-2^{14}.3^2}\)

\(=\dfrac{2^{12}.3^{14}.\left(15-14\right)}{2^{14}.\left(2.5-3\right)}\)

\(=\dfrac{3^{14}.1}{2^2}\)

\(=\dfrac{314}{4}\)

d)

\(\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}\\ =\dfrac{3^{29}.2^6.2^2}{3^{24}.3^5.2^6}\\ =\dfrac{3^{29}.2^6.4}{3^{29}.2^6}\\ =4\)

e)

\(\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}\\ =\dfrac{2^{21}.5^3.3^4}{2^3.2^{18}3^4.5}\\ =\dfrac{2^{21}.5.5^2.3^4}{2^{21}.3^4.5}\\ =5^2\\ =25\)

f)

\(=\dfrac{24\left(315+561+124\right)}{\dfrac{\left(1+99\right).50}{2}-500}\\ =\dfrac{24.1000}{2500-500}\\ =12\)

\(a,\dfrac{-14.15}{21.\left(-10\right)}=\dfrac{-7.2.3.5}{7.3.\left(-2\right).5}=1\)

\(b,\dfrac{5.7-7.9}{7.2+6.7}=\dfrac{7\left(5-9\right)}{7\left(2+6\right)}=\dfrac{-4}{8}=-\dfrac{1}{2}\)

\(c,\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}=\dfrac{-7.\left(3+4\right)}{7\left(-5-2\right)}\)

\(=\dfrac{\left(-7\right).7}{7.\left(-7\right)}=1\)

\(d,\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\dfrac{3^{29}.2^8}{3^{24}.3^5.2^6}=\dfrac{3^{29}.2^8}{3^{29}.2^6}=2^2=4\)

\(e,\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}=\dfrac{2^{21}.3^4.5^3}{2^{18}.2^3.3^4.5}=\dfrac{2^{21}.3^4.5^3}{2^{21}.3^4.5}=5^2=25\)

\(f,\dfrac{24.315+3.561.8+4.124.6}{1+3+5+...+97+99-500}\)

\(=\dfrac{24.315+24.561+24.124}{1+3+5+...+97+99-500}\)

\(=\dfrac{24\left(315+561+124\right)}{1+3+5+...+97+99-500}\)

\(=\dfrac{24.1000}{1+3+5+...+97+99-500}\) (1)

Đặt A = 1 + 3 + 5 + ... + 97 + 99

Số số hạng trong A là: (99 - 1) : 2 + 1 = 50 (số)

Tổng A bằng: (99 + 1) . 50 : 2 = 2500

Thay A = 2500 vào biểu thức (1), ta được:

\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)

a)

\(\dfrac{\left(-14\right).15}{21.\left(-10\right)}\\ =\dfrac{-7.2.3.5}{7.3.-2.5}\\=\dfrac{7.2.3.5}{7.2.3.5}\\ =1\)

b)

\(\dfrac{5.7-7.9}{7.2+6.7}\\ =\dfrac{7\left(5-9\right)}{7\left(2+6\right)}\\ =\dfrac{-4}{8}\\ =\dfrac{-2.2}{2.4}\\ =-\dfrac{1}{2}\)

c)

\(\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}\\ =\dfrac{-7.3+2.-7.2}{7\left(-5-2\right)}\\ =\dfrac{-7\left(3+4\right)}{7.-7}\\ =\dfrac{7}{7}\\ =1\)

`a) ((-14).15)/(21.(-10))`

`=(-7.3.5.3)/(7.3.(-2).5)`

`=3/2`

`b) ((-7).12)/((-16).(-14))`

`=(-7.3.2.2)/(-2.2.2.(-7).2)`

`=(-3)/4`

`c) ((-11).(-5))/(15.22) `

`=(-11.(-5))/(3.5.11.2)`

`=1/6`

`d) ((-10).(-18))/((-6).25.7)`

`=(-5.2.-2.3.3)/(-2.3.5.5.7)`

`=6/35`

`e) (5.7-7.9)/(7.2+6.7)`

`=(7.(5-9))/(7.(2+6))`

`=-4/8`

`=-1/2`

`f) (11.9-22.3)/(33.2)`

`=(33.(3-2))/(33.2)`

`=1/2`

\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)

\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)

=  \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{6}{5}\)

=  \(\dfrac{1}{4}-\dfrac{6}{5}\)

=  \(-\dfrac{19}{20}\)

\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)

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

\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)

\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)

\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)

\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)

\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)

\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)

\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)

\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)

\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)

\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)

\(=-\dfrac{5}{11}\)

\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

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

\(10\dfrac{1}{5}-5\dfrac{1}{2}\times\dfrac{60}{11}+3\div15\%\\ =\dfrac{51}{5}-\dfrac{11}{2}\times\dfrac{60}{11}+3\div\dfrac{15}{100}\\ =\dfrac{51}{5}-30+20\\ =10,2-30+20\\ =0,2\)

\(\dfrac{5}{7}\times\dfrac{5}{11}+\dfrac{5}{7}\times\dfrac{2}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times-\dfrac{7}{11}=-\dfrac{5}{11}\)

\(-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+1\dfrac{5}{7}\\ =-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times1+\dfrac{12}{7}\\ =1\)

a: \(=\dfrac{-4\cdot13\cdot9\cdot5}{3\cdot4\cdot5\cdot2\cdot13}=\dfrac{3}{2}\)

b: \(=\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot5=\dfrac{5}{6}\)

a,

\(\dfrac{4^2\cdot4^3}{2^{10}}=\dfrac{4^5}{2^{10}}=\dfrac{\left(2^2\right)^5}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)

b,

\(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{243}{0,2}=\dfrac{243}{\dfrac{1}{5}}=243\cdot5=1215\)

c,

\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^6\cdot2\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)

d,

\(\dfrac{6^3+3\cdot6^2+3^3}{-13}=\dfrac{\left(2\cdot3\right)^3+3\cdot\left(2\cdot3\right)^2+3^3}{-13}=\dfrac{2^3\cdot3^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{2^3\cdot3^3+2^2\cdot3^3+3^3}{-13}\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3\cdot13}{-13}=-3^3=-27\)

\(a,\dfrac{121.75.130.169}{39.60.11.198}=\dfrac{11.11.25.3.13.10.169}{13.3.6.10.11.11.18}=\dfrac{25.169}{6.18}\)

a) \(\dfrac{2}{3}-\dfrac{2}{6}=\dfrac{4}{6}-\dfrac{2}{6}=\dfrac{4-2}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)

b) \(\dfrac{5}{6}-\dfrac{3}{18}=\dfrac{10}{18}-\dfrac{3}{18}=\dfrac{10-3}{18}=\dfrac{7}{18}\)

c) \(\dfrac{8}{14}-\dfrac{2}{7}=\dfrac{8}{14}-\dfrac{4}{14}=\dfrac{8-4}{14}=\dfrac{4}{14}=\dfrac{2}{7}\)

d) \(\dfrac{12}{20}-\dfrac{2}{5}=\dfrac{12}{20}-\dfrac{8}{20}=\dfrac{12-8}{20}=\dfrac{4}{20}=\dfrac{1}{5}\)