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Bài 1: a) \(M=1+5+5^2+...+5^{100}\)
\(5M=5+5^2+5^3+...+5^{101}\)
\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)
\(4M=5^{101}-1\)
\(M=\frac{5^{101}-1}{4}\)
b) \(N=2+2^2+...+2^{100}\)
\(2N=2^2+2^3+...+2^{101}\)
\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(N=2^{101}-2\)
Bài 2:
a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\)
\(32^{16}=\left(2^5\right)^{16}=2^{80}\)
Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)4^50=(2^2)^50=2^100
Vậy 2^100=4^50
b) 4^3x5^3=(4x5)^3=20^3
Vì 20^3>19^3 nên 4^3x5^3>19^3
Tìm x:
3^2x4^2:(x-2)=12
(3x4)^2:(x-2)=12
12^2:(x-2)-12
x-2=12^2:12
x-2=12
x=12+2
x=14
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Trả lời:
a, Ta có: 320 ; 274 = ( 33 )4 = 312
Vì 320 > 312 nên 320 > 274
b, 225 ; 166 = ( 24 )6 = 224
Vì 225 > 224 nên 225 > 166
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
\(2^{300}=2^{3\cdot100}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=3^{2\cdot100}=\left(3^2\right)^{100}=9^{100}\)
Mà: \(8< 9\)
\(\Rightarrow8^{100}< 9^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)
b) Ta có:
\(3^{500}=3^{5\cdot100}=\left(3^5\right)^{100}=243^{100}\)
\(7^{300}=7^{3\cdot100}=\left(7^3\right)^{100}=343^{100}\)
Mà: \(243< 343\)
\(\Rightarrow243^{100}< 343^{100}\)
\(\Rightarrow3^{500}< 7^{300}\)
c) Ta có:
\(8^5=\left(2^3\right)^5=2^{3\cdot5}=2^{15}=2\cdot2^{15}\)
\(3\cdot4^7=3\cdot\left(2^2\right)^7=3\cdot2^{2\cdot7}=3\cdot2^{14}\)
Mà: \(2< 3\)
\(\Rightarrow2\cdot2^{14}< 3\cdot2^{14}\)
\(\Rightarrow8^5< 3\cdot4^7\)
d) Ta có:
\(202^{303}=202^{3\cdot101}=\left(202^3\right)^{101}=8242408^{101}\)
\(303^{202}=303^{2\cdot101}=\left(303^2\right)^{101}=91809^{101}\)
Mà: \(8242408>91809\)
\(\Rightarrow8242408^{101}>91809^{101}\)
\(\Rightarrow202^{303}>303^{202}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^{100}và3^{200}=3^{200}và3^{200}\\ \Rightarrow3^{200}=3^{200}\\ \Rightarrow9^{100}=3^{200}.\\ 5^{23}và125^3=5^{23}và5^9\\ \Rightarrow5^{23}>5^9\\ \Rightarrow5^{23}>5^3.\)
9¹⁰⁰ = (3²)¹⁰⁰ = 3²⁰⁰
Vậy 9¹⁰⁰ = 3²⁰⁰
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125³ = (5³)³ = 5⁹
Do 23 > 9 nên 5²³ > 5⁹
Vậy 5²³ > 125³
7\(^3\)và 3\(^5\)
Ta có:
7\(^3\) = 343
3\(^5\) = 243
Vì 343 > 243 nên 7\(^3\) > 3\(^5\)
Dcm