cho A= 1 + 3 +3^2+...+3^99
viết A2 + 1 dưới dạng lũy thừa cơ số 4
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ta có : \(A=1+3+3^2+...+3^{99}\)
\(\Rightarrow3A=3\left(1+3+3^2+...+3^{99}\right)\)
\(\Leftrightarrow3A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A-A=2A=\left(3+3^2+3^3+...+3^{100}\right)-\left(1+3+3^2+...+3^{99}\right)\)
\(\Leftrightarrow2A=3^{100}-1\)
\(\Rightarrow2A+1=3^{100}-1+1=3^{100}=\left(3^{25}\right)^4\)
vậy \(2A+1=\left(3^{25}\right)^4\)
Ta có: +) \({({2^2})^3} = {2^2}{.2^2}{.2^2} = {2^{2 + 2 + 2}} = {2^6}\)
+) \({\left[ {{{( - 3)}^2}} \right]^2} = {( - 3)^2}.{( - 3)^2} = {( - 3)^{2 + 2}} = {( - 3)^4}\)
\(A=1+3+3^2+...+3^{41}\)
\(3A=3+3^2+3^3+...+3^{42}\)
\(3A-A=3+3^2+...+3^{42}-1-3-...-3^{41}\)
\(2A=3^{42}-1\)
\(A=\dfrac{3^{42}-1}{2}\)
Ta có: \(2A+1\)
\(=2\cdot\dfrac{3^{42}-1}{2}+1\)
\(=3^{42}-1+1\)
\(=3^{42}\)
\(=\left(3^2\right)^{21}\)
\(=9^{21}\)
`2^5 . 8^4 = 2^5 . (2^3)^4 = 2^5 . 2^12 = 2^17`
`25^6 . 125^3 = (5^2)^6 . (5^3)^3 = 5^12 . 5^9 = 5^21`
`625^5 : 25^7 = (5^4)^7 : (5^2)^7 = 5^28 : 5^14 = 5^14`
`12^3 . 3^3 = (12 . 3)^3 = 36^3`
Ta có:
\(\begin{array}{l}{\left( {\frac{1}{4}} \right)^8} = {[{\left( {\frac{1}{2}} \right)^2}]^8} = {(\frac{1}{2})^{2.8}} = {(\frac{1}{2})^{16}};\\{\left( {\frac{1}{8}} \right)^3} = {[{(\frac{1}{2})^3}]^3} = {(\frac{1}{2})^{3.3}} = {(\frac{1}{2})^9}\end{array}\)
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}`
1.
a) \(3^4\times3^5\times3^6=3^{4+5+6}=3^{15}\)
b) \(5^2\times5^4\times5^5\times25=5^2\times5^4\times5^5\times5^2=5^{2+4+5+2}=5^{13}\)
c) \(10^8\div10^3=10^{8-3}=10^5\)
d) \(a^7\div a^2=a^{7-2}=a^5\)
2.
\(987=900+80+7\\ =9\times100+8\times10+7\\ =9\times10^2+8\times10^1+7\times10^0\)
\(2021=2000+20+1\\ =2\times1000+2\times10+1\times1\\ =2\times10^3+2\times10^1+1\times10^0\)
\(abcde=a\times10000+b\times1000+c\times100+d\times10+e\times1\\ =a\times10^4+b\times10^3+c\times10^2+d\times10^1+e\times10^0\)
A=1+3+3^2+...+3^99
3A=3+3^2+3^3+...+3^100
3A-A=3^100-1
2A=3^100-1
A=(3^100-1):2
mik chỉ làm được đến đó thôi
A=1+3+3^2+...+3^99
3A=3+3^2+3^3+...+3^100
3A-A=2A=3^100-1
\(\Rightarrow\)2A+1=3^100
Khong viet dc vi 3^100 le ma 4^n chan