Viết dưới dạng lũy thừa : 1+2+2 mũ 2 +...+2 mũ 2018
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\(A=1+2+2^2+...+2^{30}\)
\(2A=2+2^2+2^3+...+2^{31}\)
\(2A-A=\left(2+2^2+2^3+...+2^{31}\right)-\left(1+2+2^2+...+2^{30}\right)\)
\(A=2^{31}-1\)
\(A+1=2^{31}\)
\(A=1+3+3^2+...+3^{41}\)
\(3A=3+3^2+3^3+...+3^{42}\)
\(3A-A=3+3^2+...+3^{42}-1-3-...-3^{41}\)
\(2A=3^{42}-1\)
\(A=\dfrac{3^{42}-1}{2}\)
Ta có: \(2A+1\)
\(=2\cdot\dfrac{3^{42}-1}{2}+1\)
\(=3^{42}-1+1\)
\(=3^{42}\)
\(=\left(3^2\right)^{21}\)
\(=9^{21}\)
\(A=1+2+2^2+2^3+....+2^{30}\)
\(2.A=2+2^2+2^3+2^4+...+2^{30}\)
\(2.A-A=\left(2+2^2+2^3+2^4+...+2^{31}\right)-\left(1+2+2^2+2^3+...+2^{30}\right)\)
\(A=2^{31}-1\)
\(\Rightarrow A+1=2^{31}-1+1\)
\(\Rightarrow A+1=2^{31}\)
a) \(2.4.16.32.2^4=2.2^2.2^4.2^5.2^4=2^{16}\)
b) \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)=\left(2^2.2^5\right):\left(2^3.\left(\frac{1}{2}\right)^4\right)=2^7:\frac{1}{2}=2^8\)
c) \(9.3^3.\frac{1}{81}.27=3^2.3^3.\left(\frac{1}{3}\right)^4.3^3=3^4\)
d)\(2^2.4.\frac{32}{2^2}.2^5=2^2.2^2.2^3.2^5=2^{12}\)
Ta có: \(A=2+2^2+2^3+...+2^{100}\)
\(2A=2^2+2^3+2^4+...+2^{101}\)
\(2A-A=2^{101}-2\)
Hay \(A=2^{101}-2\)
Vậy \(A=2^{101}-2\)
_Học tốt_
\(13^4\cdot16^2=13\cdot13\cdot13\cdot13\cdot16\cdot16\)
\(=13\cdot13\cdot13\cdot13\cdot4\cdot4\cdot4\cdot4\)
\(=\left(13\cdot4\right)\left(13\cdot4\right)\left(13\cdot4\right)\left(13\cdot4\right)\)
\(=52\cdot52\cdot52\cdot52\)
\(=52^4\)
2^(10 : 64 x 16) = 2^[(10 x 16) : 64] = 2^(160 : 64) = 2^25
2^(10 x 8) = 2^80 3^(5 : 27) = 3^(5/27) 5^(2 x 125) = 5^250 6^(6 : 36) = 6^(1/6)