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a) Ta có :
\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)
\(\Rightarrow27^{27}>243^{13}\)
\(\Rightarrow-27^{27}< -243^{13}\)
\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)
b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)
c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)
\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)
\(\Rightarrow4^{50}>8^{30}\)
d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)
\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)
Ta có:
a) \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(5.3^2\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{5^{10}.3^{20}.5^{20}}{5^{30}.3^{15}}=\frac{5^{30}.3^{20}}{5^{30}.3^{15}}=3^5=243\)
b) \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2.2^2\right)^5}{\left(0,2.2\right)^6}=\frac{\left(0,2\right)^5.2^{10}}{\left(0,2\right)^6.2^6}=\frac{2^4}{0,2}=\frac{16}{0,2}=80\)
c) \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
a) 158 x 94
= 158 x ( 32 )4
= 158 x 38
= ( 15 x 3 )8 = 458
b) 49 : 527
= 49 : ( 53 ) 9
= 49 : 1259
= \(\left(\frac{4}{125}\right)^9\)
c) 2010 : 220
= 2010 : ( 22 )10
= 2010 : 410 = ( 20 : 4 ) 10 = 510
d) 275 : ( -7 ) 15
= 275 : [ ( - 7 )3 ]5
= 275 : ( - 21 )5
= \(\left(\frac{27}{-21}\right)^5=\left(\frac{9}{-7}\right)^5\)
Cbht
a) \(11^9+12^9+13^9+14^9+15^9+16^9\)
\(=11^{4.2}.11+12^{4.2}.12+13^{4.2}.13+14^{4.2}.14+15^9+16^9\)
\(=...1.11+...6.12+...1.13+...6.14+...5+...6\)
\(=...1+...2+...3+...4+...5+...6\)
\(=...1\)
Vậy biểu thức trên có chũ số tận cùng là 1
b) \(25^7+26^7+27^7+28^7+29^7+29^7+30^7+31^7\)
\(=...5+...6+27^4.27^3+28^4.28^3+29^4.29^3+29^4.29^3+...0+...1\)
\(=...5+...6+...3+...8+...9+...9+...0+...1\)
\(=...1\)
Vậy biểu thức trên có chữ số tận cùng là 1
a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)
b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)
c) \(27^{40}=3^{3^{40}}=3^{120}\)
\(64^{60}=8^{2^{60}}=8^{120}\)
Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)