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\(A=\frac{2^{19}.\left(2^3\right)^3+15.\left(2^2\right)^9.\left(3^2\right)^4}{2^9.3^9.2^{10}+\left(2^2.3\right)^{10}}=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^{19}.3^9+2^{20}.3^{10}}=\frac{2^{18}.3^8.\left(2.3+15\right)}{2^{19}.3^9.\left(1+2.3\right)}\)
\(=\frac{2^{18}.3^8.21}{2^{19}.3^9.7}=\frac{21}{2.3.7}=\frac{1}{2}\)
\(\frac{45^{10}.5^{10}}{75^{10}}=\frac{\left(45.5\right)^{10}}{75^{10}}=\frac{225^{10}}{75^{10}}=\left(225:75\right)^{10}=3^{10}=59049\)
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a)\(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}=\frac{3^6.\left(3^2.5\right)^4-\left(3.5\right)^{13}.5^{-9}}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}=\frac{3^6.3^8.5^4-3^{13}.5^{13}.5^{-9}}{3^{12}.5^6+3^{12}.5^6}\)
\(=\frac{3^{14}.5^4-3^{13}.5^4}{3^{12}.5^6+3^{12}.5^6}=\frac{3^{13}.5^4.\left(3-1\right)}{3^{12}.5^6\left(1+1\right)}=\frac{3^{13}.5^4}{3^{12}.5^6}=\frac{3}{5^2}=\frac{3}{25}\)
b)\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-\left(2^{12}.3^{12}+2^{11}.3^{11}\right)}=\frac{2^{12}.3^{10}\left(1+5\right)}{-\left[2^{11}.3^{11}\left(2.3+1\right)\right]}=\frac{2.6}{-\left(3.7\right)}=\frac{4}{-7}\)
\(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}=\frac{\left[2^3\right]^5\cdot\left[3^2\right]^4}{6^6\cdot\left[2^3\right]^3}=\frac{2^{15}\cdot3^8}{6^6\cdot2^9}=\frac{2^6\cdot3^8}{6^6}=\frac{\left[2^2\right]^3\cdot\left[3^2\right]^4}{\left[2\cdot3\right]^6}=\frac{2^6\cdot3^8}{2^6\cdot3^6}=\frac{3^8}{3^6}=3^2=9\)
Câu 1
a) (3,1-2,5) -( -2,5 +3,1) = 3,1 -2,5 +2,5 - 3,1 =0
b) 253 : 52 = (52)3 : 52 = 56 : 52 =54 = 625
c) 0,254 . 1024 = 1/256 . 1024 = 4
d)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^{15}3^6}=3^2\)\(=9\)
Câu 2
a)\(x-\frac{3}{4}=\frac{2}{3}\)=> \(x=\frac{2}{3}+\frac{3}{4}=\frac{17}{12}\)
Vậy x\(=\frac{17}{12}\)
b) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)=>\(\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)=>\(\frac{31}{60}-x=\frac{2}{3}\)=>\(x=\frac{31}{60}-\frac{2}{3}=-\frac{3}{20}\)
\(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}=\frac{2^{15}\cdot3^4\cdot3^4}{2^6\cdot3^6\cdot\left(2^4\right)^3}=\frac{2^{12}\cdot2^3\cdot3^8}{2^6\cdot3^6\cdot2^{12}}\)
\(=\frac{2^3\cdot3^2\cdot3^6}{2^3\cdot2^3\cdot3^6}=\frac{3^3}{2^3}\)
\(=\frac{27}{8}\)
\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{\left(2^3\right)^5.\left(3^2\right)^4}{6^6.8^3}=\frac{8^5.3^8}{6^6.8^3}\)
\(=\frac{8^2.3^8}{6^6}=\frac{64.6561}{46656}=\frac{64.6561}{64.729}\)
\(=\frac{6561}{729}=9\)
\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=2.3^2=2.9=18\)
\(C=\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{3^6.2^6.2^9}=\frac{2^{15}.3^8}{3^6.2^{15}}=3^2=9\)
C=\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^8}{3^6.2^{15}}=3^2\)\(=9\)
\(\frac{2^{15}.9^4}{6^6.8^3}\)
\(=\)\(\frac{2^{15}.\left(3^2\right)^4}{\left(3.2\right)^6.\left(2^3\right)^3}\)
\(=\)\(\frac{2^{15}.3^8}{3^6.2^6.2^9}\)
\(=\)\(\frac{2^{15}.3^6.3^2}{3^6.2^{15}}\)
\(=\)\(3^2=9\)