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a) 9.33.1/81.32=32.33.3-4.32=33
b)4.25:(23. 1/16)=22.25:(23.2-4)=27:2-1=27:(1/2)=27.2=28
c) 32.25.(2/3)2=32.25.22/32=27
d) (1/3)2.1/3.92=3-2.3-1.34=31
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, 0,5^3 . 4^3 : ( 16 . 1/8) = 0,5^3 . 4^3 : 2 =(0,5.4)^3 : 2 = 2^3 :2 = 2^2
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)\(3^4.3^5:\frac{1}{27}=3^9:\frac{1}{3^3}=3^9.3^3=3^{12}\)
\(A=1+2+2^2+2^3+...+2^{30}\)
\(2A=2+2^2+2^3+...+2^{31}\)
\(2A-A=2+2^2+2^3+...+2^{31}-1-2-2^2-2^3-...-2^{30}\)
\(A=2^{31}-1\)
\(A=1+2+2^2+2^3+...+2^{30}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{31}\)
\(\Rightarrow2A-A=A=2^{31}-1\Leftrightarrow A+1=2^{31}-1+1=2^{31}\)
Vậy \(A+1=2^{31}\)