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Ta có: A = 1 + 2 + 22 + 23 + ....... + 2200
=> 2A = 2 + 22 + 23 + ....... + 2201
=> 2A - A = ( 2 + 22 + 23 + ....... + 2201 ) - ( 1 + 2 + 22 + 23 + ....... + 2200 )
=> A = 2201 - 1
=> A + 1 = 2201
A = 1 + 2 + 2 ^ 2 + 2 ^ 3 + ... + 2 ^ 200
2A = 2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + ... + 2 ^ 201
2A - A = ( 2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + ... + 2 ^ 201 )
- ( 1 + 2 + 2 ^ 2 + 2 ^ 3 + ... + 2 ^ 200 )
A = 2 ^ 201 - 1
=> A + 1 = 2 ^ 201
B = 3 + 3 ^ 2 + 3 ^ 3 + ... + 3 ^ 2005
3B = 3 ^ 2 + 3 ^ 3 + 3 ^ 4 + ... + 3 ^ 2006
3B - B = ( 3 ^ 2 + 3 ^ 3 + 3 ^ 4 + ... + 3 ^ 2006 )
- ( 3 + 3 ^ 2 + 3 ^ 3 + ... + 3 ^ 2005 )
2B = 3 ^ 2006 - 3
=> 2B = 3 ^ 2006
Vậy 2B + 3 là lũy thừa của 3
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`@` `\text {Ans}`
`\downarrow`
\(3^2\cdot2^5\cdot\left(\dfrac{2}{3}\right)^2\)
`=`\(\left(3\cdot\dfrac{2}{3}\right)^2\cdot2^5\)
`=`\(2^2\cdot2^5=2^7\)
\(3^2\cdot2^5\cdot\left(\dfrac{2}{3}\right)^2\)
\(=2^5\cdot\left(3\cdot\dfrac{2}{3}\right)^2\)
\(=2^5\cdot\left(\dfrac{3\cdot2}{3}\right)^2\)
\(=2^5\cdot2^2\)
\(=2^{2+5}\)
\(=2^5\)
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b: \(3^4\cdot3^5:\dfrac{1}{27}==3^9\cdot3^3=3^{12}\)
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\(A=1+2+2^2+2^3+...+2^{30}\)
\(2A=2+2^2+2^3+...+2^{31}\)
\(2A-A=2+2^2+2^3+...+2^{31}-1-2-2^2-2^3-...-2^{30}\)
\(A=2^{31}-1\)
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Ta có: +) \({({2^2})^3} = {2^2}{.2^2}{.2^2} = {2^{2 + 2 + 2}} = {2^6}\)
+) \({\left[ {{{( - 3)}^2}} \right]^2} = {( - 3)^2}.{( - 3)^2} = {( - 3)^{2 + 2}} = {( - 3)^4}\)
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\(\left(-\frac{3}{4}\right)^{10}:\left(-\frac{1}{2}\right)^{10}\)
\(=\left(\frac{-3}{4}:\frac{-1}{2}\right)^{10}\)
\(=\left(\frac{3}{2}\right)^{10}\)
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\(27^3:9\left[\left(-3\right)^2\right]^2=3^9:3^2\cdot3^4=3^{9-2+4}=3^{11}\)
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a) 8 = 23
425 = 25.35.75
16 = 24
b) (0,09)3 = (3/10)6
(3/10)8 = (3/10)8
0,027 = (3/10)3
`@` `\text {Ans}`
`\downarrow`
`a)`
`8 = 2^3`
`32^5` chứ ạ?
`32^5 = (2^5)^5 = 2^10`
`16 = 2^4`
`b)`
`(0,09)^3 = (0,3^2)^3 = 0,3^6` hay `(3/10)^6`
`(3/10)^8 = (3/10)^8`
`(0,027) = (0,3)^3` hay `(3/10)^3`
`@` `\text {Kaizuu lv uuu}`
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