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\(5^{45}=5.\left(5^2\right)^{22}=5.25^{22}\)
\(3^{73}=3^4.\left(3^3\right)^{23}=81.27^{23}\)
\(\Rightarrow5^{45}< 3^{73}\)
\(2^{83}=2^2.\left(2^3\right)^{27}=4.8^{27}\)
\(3^{57}=3^3.\left(3^2\right)^{27}=27.9^{27}\)
\(\Rightarrow2^{83}< 3^{57}\)
a) \(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
vì 8 < 9 và 75 = 75
=> 875 < 975
=> 2225 < 3150
b) \(2^{91}>2^{90}=\left(2^5\right)^{18}=32^{18}>25^{18}=5^{36}>5^{35}\)
\(\Rightarrow2^{91}>5^{35}\)
c) \(5^{300}=\left(5^3\right)^{100}=125^{100}\)
\(3^{500}=\left(3^5\right)^{100}=243^{100}\)
Vì 125 < 243 mà 100 = 100
=> \(5^{300}< 3^{500}\)
Bài nì lp 6 lm nhìu rùi mà
Ta có:
+ 2225 = (23)75 = 875
3150 = (32)75 = 975
Vì 875 < 975
=> 3225 < 3150
+ 291 = (213)7 = 81927
535 = (55)7 = 31257
Vì 81927 > 31257
=> 291 > 535
+ 5300 = (53)100 = 125100
3500 = (35)100 = 243100
Vì 125100 < 243100
=> 5300 < 3500
bài 2
làm câu B;C nha
B)
\(27^3=\left(3^3\right)^3=3^9\)
\(9^5=\left(3^2\right)^5=3^{10}\)
vì \(10>9\)
\(=>9^5>27^3\)
C)
\(\left(\frac{1}{8}\right)^6=\left(\frac{1}{2^3}\right)^6=\frac{1^6}{2^{18}}=\frac{1}{2^{18}}\)
\(\left(\frac{1}{32}\right)^4=\left(\frac{1}{2^5}\right)^4=\frac{1^4}{2^{20}}=\frac{1}{2^{20}}\)
vì \(2^{18}< 2^{20}\)
\(=>\frac{1}{2^{18}}>\frac{1}{2^{20}}\)
\(=>\left(\frac{1}{8}\right)^6>\left(\frac{1}{32}\right)^4\)
\(\text{A.}\frac{32^3.9^5}{8^3.6^6}=\frac{\left(2^5\right)^3.\left(3^2\right)^5}{\left(2^3\right)^3.\left(2.3\right)^6}=\frac{2^{15}.3^{10}}{2^9.2^6.3^6}=\frac{3^{10}}{3^6}=3^4=81\)
\(\text{B.}\frac{\left(5^5-5^4\right)^3}{50^6}=\frac{2500^3}{50^6}=\frac{\left(50^2\right)^3}{50^6}=\frac{50^6}{50^6}=1\)
Bài 2:
\(\text{A.Ta có:}\)
\(5^6=\left(5^3\right)^2=125^2\)
\(\left(-2\right)^{14}=2^{14}=\left(2^7\right)^2=128^2\)
Vì \(125< 128\)
\(\Rightarrow125^2< 128^2\)
\(\Rightarrow5^6< \left(-2\right)^{14}\)
\(\text{B.Ta có:}\)
\(9^5=\left(3^2\right)^5=3^{10}\)
\(27^3=\left(3^3\right)^3=3^9\)
Vì \(9< 10\)
\(\Rightarrow3^9< 3^{10}\)
\(\Rightarrow27^3< 9^5\)
\(\text{C.Ta có:}\)
\(\left(\frac{1}{8}\right)^6=\left[\left(\frac{1}{2}\right)^3\right]^6=\left(\frac{1}{2}\right)^{18}\)
\(\left(\frac{1}{32}\right)^4=\left[\left(\frac{1}{2}\right)^5\right]^4=\left(\frac{1}{2}\right)^{20}\)
Vì \(18< 20\)
\(\Rightarrow\left(\frac{1}{2}\right)^{18}< \left(\frac{1}{2}\right)^{20}\)
\(\Rightarrow\left(\frac{1}{8}\right)^6< \left(\frac{1}{32}\right)^4\)
Xét: \(\frac{\left(17^{2017}+16^{2017}\right)^{2018}}{17^{2017.2018}}=\left(\frac{17^{2017}+16^{2017}}{17^{2017}}\right)^{2018}=\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}\)
\(\frac{\left(17^{2018}+16^{2018}\right)^{2017}}{17^{2017.2018}}=\left(\frac{17^{2018}+16^{2018}}{17^{2018}}\right)^{2017}=\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
Ta có: \(0< \frac{16}{17}< 1\)
=> \(\left(\frac{16}{17}\right)^{2017}>\left(\frac{16}{17}\right)^{2018}\)
=> \(1+\left(\frac{16}{17}\right)^{2017}>1+\left(\frac{16}{17}\right)^{2018}>1\)
=> \(\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}>\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
=> \(\left(17^{2017}+16^{2017}\right)^{2018}>\left(17^{2018}+16^{2018}\right)^{2017}\)
Ta sẽ so sánh \(5^{199}\) và \(3^{300}\)
Mà:\(5^{199}< 5^{200}=25^{100}< 27^{100}=3^{300}\)
\(\Rightarrow5^{199}< 3^{300}\Rightarrow\frac{1}{5^{199}}>\frac{1}{3^{300}}\)
2/3<1 nên lũy thừa càng cao càng nhỏ
(2/3)300<2/3<1
(3/2)200>3/2>1
Ta có 3451 > 3450
Lại có 5300 = ( 52 )150 = 25150
3450 = ( 33 )150 = 27150
Vậy 25150 < 27150 hay 5300 < 3450 mà 3451 > 3450 nên 5300 < 3451
Ta có:
\(5^{300}=\left(5^2\right)^{150}=25^{150}\\ 3^{453}>3^{450}=\left(3^3\right)^{150}=27^{150}\\ Vì...27^{150}>25^{150}\Rightarrow3^{453}>3^{450}>5^{300}\\ \Rightarrow3^{453}>5^{300}\)