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1:
a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)
b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)
nên 2 p/s ko bằng nhau
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1) \(15^{40}=15^{4.10}=\left(15^4\right)^{10}=50625^{10}\)
\(18^{30}=18^{3.10}=\left(18^3\right)^{10}=5832^{10}\)
Vì \(50625^{10}>5832^{10}\) Nên \(15^{40}>18^{30}\)
2)\(180^{300}=180^{3.100}=\left(18^3\right)^{100}=5832^{100}\)
Vì \(2015^{100}
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà 17^19+1>17^18+1
nên A<B
b: \(2C=\dfrac{2^{2021}-2}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)
\(2D=\dfrac{2^{2022}-2}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)
2^2021-1<2^2022-1
=>1/2^2021-1>1/2^2022-1
=>-1/2^2021-1<-1/2^2022-1
=>C<D
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a) 1619 và 825
Ta có :
1619 = ( 24 )19 = 276
825 = ( 23 )25 = 275
Vì 276 > 275 Nên 1619 > 825
b) 536 và 1124
Ta có :
536 = ( 53 )12 = 12512
1124 = ( 112 )12 = 12112
Vì 12512 > 12112 Nên 536 > 1124
1.
\(M=3^0+3^1+......+3^{50}.\)
\(\Rightarrow3M=3+3^2+.......+3^{51}\)
\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)
\(\Rightarrow2M=3^{51}-1\)
\(\Rightarrow M=\frac{3^{51}-1}{2}\)
2.
\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)
\(8^{25}=\left(2^3\right)^5=2^{75}\)
Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)
\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)