\(C=\dfrac{2^6\cdot3^{10}}{3^9\cdot2^6}=3\\ D=\dfrac{3^{24}\cdot3^{10}}{3^{21}\cdot3^{11}}=\dfrac{3^{34}}{3^{32}}=3^2=9\\ F=\dfrac{2^{45}\cdot5^{14}}{5^{15}\cdot2^{47}}=\dfrac{1}{2^2\cdot5}=\dfrac{1}{20}\\ G=\dfrac{2^2\cdot5^2\cdot5^3}{2^3\cdot5^4}=\dfrac{1\cdot5}{2}=\dfrac{5}{2}\)

C=3

D=9

F=1/20

G=5/2

Em ko giải chi tiết vì nó lâu

Mong thông cảm!

a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\)

\(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\)

\(=\dfrac{18}{5}+\dfrac{2}{3}+4\)

\(=\dfrac{124}{15}\)

b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\)

\(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\)

\(=-\dfrac{71}{375}\)

c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\)

\(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\)

=1+2/5

=7/5

d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\)

e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)

Câu 1 :

\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)

\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)

\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)

Câu 2 :

\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)

Sorry . Mình chỉ biết đến đây thôi

a,\(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)

\(\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(x=-\dfrac{13}{15}\)

b,\(\left(3x+1\right)^3=-27\)

\(3x+1=-3\)

\(3x=-4\)

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

a ) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\\\dfrac{1}{5}-\dfrac{3}{2}x=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=-\dfrac{13}{10}\\\dfrac{3}{2}x=\dfrac{17}{10}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{15}\\x=\dfrac{17}{15}\end{matrix}\right.\)

Vậy .......

b ) \(\left(3x+1\right)^3=-27\)

\(\Leftrightarrow3x+1=-3\)

\(\Leftrightarrow3x=-4\Leftrightarrow x=-\dfrac{4}{3}\)

Vậy ..........

c ) \(7.3^{x-1}-3^{x+2}=-540\)

\(\Leftrightarrow7.3^{x-1}-3^{x-1}.3^3=-540\)

\(\Leftrightarrow3^{x-1}\left(7-27\right)=-540\)

\(\Leftrightarrow-20.3^{x-1}=-540\)

\(\Leftrightarrow3^{x-1}=27\)

\(\Leftrightarrow x=4\).

Vậy ..........

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)

\(=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{12}{15}=\dfrac{4}{5}\)

\(\dfrac{4^6.9^5+6^9.120}{8^4.3^2-6^{11}}\)

\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.120}{\left(2^3\right)^4.3^2-\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^2-2^{11}.3^{11}}\)

\(=\dfrac{1.3^8+1.1.120}{1.1-4.9}\)

\(=\dfrac{1.6561+120}{1-36}=\dfrac{6681}{-35}\)

Chắc sai đấy!!!!

\(\dfrac{4^6.9^5+6^9.120}{8^4.3^2-6^{11}}\)

\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^2-\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^2-2^{11}.3^{11}}\)

\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^2-2^{11}.3^{11}}\)

\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^2\left(2-3^9\right)}\)

\(=\dfrac{2.3^8.6}{2-3^9}=\dfrac{2.3^8.2.3}{2-3^9}=\dfrac{2^2.3^9}{2-3^9}\)

ta có : \(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\dfrac{4\left(4^5.9^5+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}=\dfrac{4\left(2^{10}.3^{10}+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}\)

\(=\dfrac{4\left(6^{10}+5.6^{10}\right)}{-6^{12}-6^{11}}=\dfrac{4.6^{11}}{-6^{11}\left(6+1\right)}=-\dfrac{4}{7}\)

\(=\dfrac{4}{5}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{-2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=-\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}=\dfrac{-2}{3}\)