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a/
+ \(\frac{1}{243^6}=\frac{1}{3^6.81^6}=\frac{1}{3^2.3^4.81^6}=\frac{1}{9.81^7}\) (1)
+ \(80< 81\Rightarrow80^7< 81^7\Rightarrow\frac{1}{80^7}>\frac{1}{81^7}\) (2)
+ \(81^7< 9.81^7\Rightarrow\frac{1}{81^7}>\frac{1}{9.81^7}\) (3)
Từ (1) (2) (3) \(\Rightarrow\frac{1}{80^7}>\frac{1}{243^6}\)
b/ Xem lại đề bài
Bài 1 :
a)25^4 và 125^3
Ta có
25^4=(5^2)^4=5^8
125^3=(5^3)^3=5^9
Vì 5^9 > 5^8 nên 125^3 > 25^4
b) \(6^{38}>6^{37}>5^{37}\)
c) \(243^4=3^{20};27^6=3^{18}< 3^{20}\)
d) \(7^{48}>7^{72}>6^{72}\)
275 =(33)5= 315
2433 x 36 = (35)3 x 36 =315 x 36
Vì 315<315 x 36 nên 275 <2433 x 36
a, Ta có: \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7=\left(\dfrac{1}{3^4}\right)^7=\left(\dfrac{1}{3}\right)^{28}=\dfrac{1}{3^{28}}\)
\(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3^5}\right)^6=\left(\dfrac{1}{3}\right)^{30}=\dfrac{1}{3^{30}}\)
Vì \(\dfrac{1}{3^{28}}>\dfrac{!}{3^{30}}\Rightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\Rightarrow\) \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)
b, Ta có: \(\left(\dfrac{3}{8}\right)^5=\dfrac{3^5}{\left(2^3\right)^5}=\dfrac{243}{2^{15}}>\dfrac{243}{3^{15}}>\dfrac{125}{3^{15}}=\dfrac{5^3}{\left(3^5\right)^3}=\left(\dfrac{5}{243}\right)^3\)
\(\Rightarrow\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)
Ta có :
\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)7=\frac{1}{3^{28}}\)
\(\left(\frac{1}{213}\right)^6>\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)
Có : (1/80)^7 < (1/64)^7 = [(1/2)^6]^7 = (1/2)^42
(1/213)^6 > (1/218)^6 = [(1/2)^7]^6 = (1/2)^42
=> (1/80)^7 < (1/213)^6
Tk mk nha
Có:2437=(35)7=335
910x275=(32)10x(33)10=320x330=350
=>..
Ta có:910 = (32)10 =320
275 = (33)5 =315
2437 =(35)7=335
910+275=320+315=335
Mà 335=335
Vậy 2437=910+275
\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\) nên \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
Ta có : \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1}{\left(3 ^4\right)^7}=\frac{1}{3^{28}}\)(1)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{^{\left(3^5\right)^6}}=\frac{1}{3^{30}}<\frac{1}{3^{28}}\)(2)
Từ (1) và (2) => \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)