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a)
3210 = (25)10 = 250
1615 = (24)15 = 260
Vì 50 < nên 250 < 260
Vậy 3210 < 1615
b)
2711 = (33)11 = 333
818 = (34)8 = 332
Vì 33 > 32 nên 333 > 332
Vậy 2711 > 818
c)
536 = (53)12 = 12512
1124 = (112)12 = 12112
Vì 125 > 121 nên 12512 > 12112
Vậy 536 > 1124
d)
216 = 23.213 = 8.213
Vì 7 < 8 nên 7.213 < 8.213
Vậy
a) Ta có: 3210 = (25)10 = 250
1615 = (24)15 = 260
Vì 50 < 60 => 250 < 260 => 3210 < 1615
b) Ta có: 2711 = (33)11 = 333
818 = (34)8 = 332
Vì 33 > 32 => 333 > 332 => 2711 > 818
c) Ta có: 536 = (53)12 = 12512
1124 = (112)12 = 12112
Vì 125 > 121 => 12512 > 12112 => 536 > 1124
d) Ta có: 216 = 213.23 = 8.213
Vì 7 < 8 => 7.213 < 8.213 => 7.213 < 8.213
a) 410.220
= 410.(22)10
= 410.410
= (4.4)10
= 1610
a) \(\dfrac{15^{30}}{45^{15}}=\dfrac{15^{30}}{3^{15}.15^{15}}=\dfrac{15^{15}}{3^{15}}=5^{15}\)
b) \(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{8^5.3^8}{2^6.3^6.8^3}=\dfrac{8^2.3^2}{2^6}=\dfrac{2^6.3^2}{2^6}=3^2=9\)
c) \(\dfrac{14^{10}.21^{32}.35^{48}}{10^{10}.15^{32}.7^{96}}=\dfrac{2^{10}.7^{10}.3^{32}.7^{32}.5^{48}.7^{48}}{2^{10}.5^{10}.3^{32}.5^{32}.7^{96}}\)
= \(\dfrac{2^{10}.7^{58}.3^{32}.5^{48}}{2^{10}.5^{42}.3^{32}.7^{96}}=\dfrac{5^6}{7^{38}}\) ( Câu này làm bừa, có lẽ sai đấy :)) )
2. So sánh
a) 3200 = 9100
2300 = 8100
Vì 9100 > 8100 nên 3200 < 2300
b) 912 = 7294
268 = 6764
Vì 7294 > 6764 nên 912 > 268
c) 224 = 88
316 = 98
Vì 88 < 98 nên 224 < 316
\(1.a)\) Ta có: \(\left\{{}\begin{matrix}64^8=\left(8^2\right)^8=8^{16}\\16^{12}=8^{12}.2^{12}=8^{12}.\left(2^3\right)^4=8^{12}.8^4=8^{16}\end{matrix}\right.\)
Có: \(8^{16}=8^{16}\Rightarrow64^8=16^{12}\)
Vậy...
\(b)\) Ta có: \(\left\{{}\begin{matrix}\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\\\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\end{matrix}\right.\)
Có: \(\left(-125\right)^{10}< \left(-243\right)^{10}\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
Vậy...
\(c)\) Ta có: \(\left\{{}\begin{matrix}2^{27}=\left(2^3\right)^9=8^9\\3^{18}=\left(3^2\right)^9=9^9\end{matrix}\right.\)
Có: \(8^9< 9^9\Rightarrow2^{27}< 3^{18}\)
Vậy...
\(d)\) Ta có: \(\left\{{}\begin{matrix}\left(\dfrac{1}{25}\right)^{10}=\left[\left(\dfrac{1}{5}\right)^2\right]^{10}=\left(\dfrac{1}{5}\right)^{20}\\\left(\dfrac{1}{125}\right)^8=\left[\left(\dfrac{1}{5}\right)^3\right]^8=\left(\dfrac{1}{5}\right)^{24}\end{matrix}\right.\)
Có: \(\left(\dfrac{1}{5}\right)^{20}< \left(\dfrac{1}{5}\right)^{24}\Rightarrow\left(\dfrac{1}{24}\right)^{10}< \left(\dfrac{1}{125}\right)^8\)
Vậy...
\(e)\)Có: \(32^9=\left(2^5\right)^9=2^{45}< 2^{52}=\left(2^4\right)^{13}=16^{13}< 18^{13}\)
\(\Rightarrow32^9< 18^{13}\)
Vậy...
\(\frac{16^{12}.5^{30}.125^6}{32^{10}.81^7.25^{23}}=\frac{2^{48}.3^{30}.5^{30}.5^{18}}{2^{50}.3^{28}.5^{46}}\)
\(=\frac{2^{48}.3^{28}.3^2.5^{46}.5^2}{2^{48}.2^2.3^{28}.5^{46}}\)
=\(\frac{3^2.5^2}{2^2}\)