13)\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}\right)+\dfrac{45}{8}\left(\dfrac{11}{15}+\dfrac{9}{8}\right)\)

=\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}+\dfrac{11}{15}+\dfrac{9}{8}\right)\)

=\(\dfrac{45}{8}\left[\left(\dfrac{4}{15}+\dfrac{11}{15}\right)-\left(\dfrac{7}{8}-\dfrac{9}{8}\right)\right]\)

=\(\dfrac{45}{8}.\dfrac{5}{4}\)=\(\dfrac{225}{32}\)

14)\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right):\dfrac{7}{15}\)

=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right).\dfrac{15}{7}\)

=\(\dfrac{15}{7}\left[\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right)\right]\)

=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}-\dfrac{2}{5}+\dfrac{27}{4}\right)\)

=\(\dfrac{15}{7}.\dfrac{35}{8}\)=\(\dfrac{75}{8}\)

ai giúp mìn vứi ❤

a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)

b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)

c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)

\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)

a) \(\dfrac{3^6\cdot3^8\cdot5^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+3^{12}\cdot5^6}=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{3^{12}\cdot5^6\cdot2}=\dfrac{3^{12}\cdot5^4\left(3^2-3\right)}{3^{12}\cdot5^6\cdot2}=\dfrac{3^2-3}{5^2\cdot2}=\dfrac{6}{50}=\dfrac{3}{25}\)

a: \(=\dfrac{-3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)=\dfrac{-3}{4}\cdot40=-30\)

b: \(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)

\(=\dfrac{35}{6}:\dfrac{-175+132}{42}+\dfrac{15}{2}\)

\(=\dfrac{35}{6}\cdot\dfrac{42}{-43}+\dfrac{15}{2}\)

\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}\)

\(=\dfrac{-70\cdot7+15\cdot43}{86}=\dfrac{155}{86}\)

c: \(=\dfrac{-7}{5}\left(4+\dfrac{5}{9}+5+\dfrac{4}{9}\right)=\dfrac{-7}{5}\cdot10=-14\)

d: \(=4+\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}\cdot\dfrac{64}{125}\cdot\dfrac{-8}{27}\right)\)

\(=\dfrac{89}{16}+25\cdot\dfrac{-32}{375}\)

\(=\dfrac{89}{16}-\dfrac{32}{15}=\dfrac{823}{240}\)

e: \(=\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)=\dfrac{2}{3}-5=-\dfrac{13}{3}\)

3)\(\dfrac{-41}{32}\left(\dfrac{15}{8}-\dfrac{16}{41}\right)+\dfrac{15}{8}\left(\dfrac{41}{32}-\dfrac{8}{3}\right)\)

=\(\dfrac{-41}{32}.\dfrac{15}{8}-\dfrac{-41}{32}.\dfrac{16}{41}+\dfrac{15}{8}.\dfrac{41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)

=\(\left(\dfrac{-41}{32}.\dfrac{15}{8}+\dfrac{15}{8}.\dfrac{41}{32}\right)+\dfrac{-16}{41}.\dfrac{-41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)

=\(0+\dfrac{1}{2}-5=\dfrac{-9}{2}\)

4)\(\dfrac{13}{29}\left(\dfrac{29}{5}-\dfrac{45}{8}\right)-\dfrac{45}{8}\left(\dfrac{9}{8}-\dfrac{13}{29}\right)\)

=\(\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{9}{8}-\dfrac{45}{8}.\dfrac{13}{29}\)

=\(\left(\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{13}{29}\right)-\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{9}{8}\)

=\(0-\dfrac{13}{5}-\dfrac{405}{64}=\dfrac{-2857}{320}\)

Bài 1:

a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)

b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{10}\)

c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)

\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)

\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)

\(=1\)

Bài 2:

a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)

b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)

c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)

d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)

Bài 3:

a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)

\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)

\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)

\(=9.\dfrac{2}{3}.-1\)

\(=6.-1=-6\)

b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)

\(#Wendy.Dang\)

Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))

Võ Ngọc Phương

Bài  3b, kết quả -(3)⁶ = - 729 em nhá chứ không phải (-3)⁶

a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)

\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)

\(=-1+1=0\)

b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)

\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)

=1-1+1=1

7: \(=\dfrac{-12}{7}\cdot15+\dfrac{2}{7}\cdot\left(-15\right)+\left(-105\right)\cdot\dfrac{70-84+15}{105}\)

\(=\dfrac{-12\cdot15+2\cdot\left(-15\right)}{7}-1\)

\(=\dfrac{-15\cdot14}{7}-1=-15\cdot2-1=-31\)

8: \(=\dfrac{13}{29}\cdot\dfrac{29}{5}-\dfrac{13}{29}\cdot\dfrac{45}{8}-\dfrac{45}{8}\cdot\dfrac{9}{8}+\dfrac{45}{8}\cdot\dfrac{13}{29}\)

\(=-\dfrac{1193}{320}\)