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ta có : \(4^3.3^7.12^6.9^3.8^2=\left(2^2\right)^3.3^7.\left(2^2.3\right)^6.\left(3^2\right)^3.\left(2^3\right)^2\)
= \(2^6.3^7.2^{12}.3^6.3^6.2^6=2^{24}.3^{19}\)
\(4^3.3^7.12^6.9^3.8^2=2^6.3^7.2^{12}.3^6.3^6.2^6=2^{24}.3^{19}\)
Đề này là chịu rồi
\(4^3\cdot3^7\cdot12^6\cdot9^3\cdot8^2=2^6\cdot3^7\cdot2^{12}\cdot3^6\cdot3^6\cdot2^6\)
\(=2^{24}\cdot3^{19}\)
...........đến đây chịu!!....
a, 0,5^3 . 4^3 : ( 16 . 1/8) = 0,5^3 . 4^3 : 2 =(0,5.4)^3 : 2 = 2^3 :2 = 2^2
+) \(1,5^4\times3^4=\left(1,5\times3\right)^4=4,5^4\)
+) \(\left(-5\right)^3\times\left(-5\right)^2=\left(-5\right)^{3+2}=\left(-5\right)^5\)
+) \(12^3:4^3=\left(12:4\right)^3=3^3\)
+) \(3^{15}:3^{12}=3^{15-12}=3^3\)
_Chúc bạn học tốt_
A)4x2^5:(2^3.1/16)=2^2.2^5:(8.1/16)=2^7:1/2=2^7.2=2^8
B)3^2.2^5.(2/3)^2=3^2.2^5.4/9=2^5.9.4/9=2^5.4=2^5.2^2=2^7
C) 9.3^3.1/81.3^2=3^3.9.9.1/81=3^3.81.1/81=3^3
a) 4x32:(\(^{2^3}\)x\(\frac{1}{16}\))=\(^{2^2}\)x\(2^5\):[\(2^3\)x\(\left(\frac{1}{2}\right)^4\)] =\(2^7\):(\(2^3\)x\(2^{-4}\))=\(2^7\):\(2^{-1}\)=\(2^{7-\left(-1\right)}\)=\(2^8\)
b)\(3^4\)x\(3^5\):\(\frac{1}{27}\)=\(3^9\):\(\left(\frac{1}{3}\right)^3\)=\(3^9\):\(3^{-3}\)=\(3^{9-\left(-3\right)}\)=\(3^{12}\)
a)\(4.32:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^7.2=2^8\)
b)\(3^4.3^5:\frac{1}{27}=3^9:\frac{1}{3^3}=3^9.3^3=3^{12}\)
\(A=1+2+2^2+2^3+...+2^{30}\)
\(2A=2+2^2+2^3+...+2^{31}\)
\(2A-A=2+2^2+2^3+...+2^{31}-1-2-2^2-2^3-...-2^{30}\)
\(A=2^{31}-1\)
\(2^3.4^5=8.1024=2^3.2^{10}=2^{13}\)