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(\(\frac{1}{2}\))50=(\(\frac{1}{2^5}\))10=(\(\frac{1}{32}\))10
Do 1/6> 1/30 nên (\(\frac{1}{6}\))10>(\(\frac{1}{2}\))50
\(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left[\frac{1^5}{2^5}\right]^{10}=\left[\frac{1}{32}\right]^{10}\)
Vì 2 phân số này có cùng tử mà 6 < 30
=> \(\frac{1}{6}>\frac{1}{30}\)
=> \(\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
1,1020và 9010
ta có:+,1020=(102)10=10010
+,9010=9010
vì 10010>9010=>1020>9010
2,(1/16)10 và (1/2)50
ta có:+, (1/16)10=(1/16)10
+,(1/2)50=(1/25)10=(1/32)10
vì (1/16)10>(1/32)10=>(1/16)10>(1/2)50
k mik nhé
\(a,\) \(10^{20}=10^{10+10}=10^{10}.10^{10}\)
\(90^{10}=9^{10}.10^{10}\)
Vì \(10^{10}.10^{10}>9^{10}.10^{10}\)
\(\Rightarrow10^{20}>90^{10}\)
Vậy \(10^{20}>90^{10}\)
\(b,\)\(\left(\frac{1}{16}\right)^{10}=\frac{1^{10}}{16^{10}}=\frac{1}{\left(4^2\right)^{10}}=\frac{1}{4^{20}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1^{50}}{2^{50}}=\frac{1}{\left(2^2\right)^{25}}=\frac{1}{4^{25}}\)
Vì \(\frac{1}{4^{20}}>\frac{1}{4^{25}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
~~~~~~~~~~Hok tốt~~~~~~~~~~~
\(\left(\dfrac{1}{2}\right)^{50}=\left[\left(\dfrac{1}{2}\right)^5\right]^{10}=\left(\dfrac{1}{32}\right)^{10}\)
1/12>1/32
=>(1/12)^10>(1/32)^10
=>(1/12)^10>(1/2)^50
Có: \(\left(\dfrac{1}{12}\right)^{10}=\dfrac{1}{12^{10}}\)
\(\left(\dfrac{1}{2}\right)^{50}=\dfrac{1}{2^{50}}=\dfrac{1}{\left(2^5\right)^{10}}=\dfrac{1}{32^{10}}\)
Do \(12< 32\Rightarrow12^{10}< 32^{10}\)
\(\Rightarrow\dfrac{1}{12^{10}}>\dfrac{1}{32^{10}}\) hay \(\left(\dfrac{1}{12}\right)^{10}>\left(\dfrac{1}{2}\right)^{50}\)
\(\left(\frac{1}{16}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)
vì 40<50 nên \(\left(\frac{1}{2}\right)^{40}
\(\left(\dfrac{7}{2}\right)^{50}=\left(\dfrac{16807}{32}\right)^{10}\)
mà 16807/32>1/16
nên \(\left(\dfrac{1}{16}\right)^{10}< \left(\dfrac{7}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
Vì \(2^{40}< 2^{50}\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)hay \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(0,3\right)^{20}=\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}\)
Vì \(0,09< 0,1\Rightarrow\left(0,09\right)^{10}< \left(0,1\right)^{100}\)
hay \(\left(0,3\right)^{20}< \left(0,1\right)^{10}\)
1) \(5^{199}< 5^{200}=25^{100}\)
\(3^{300}=27^{100}>25^{100}\)
\(\Rightarrow3^{300}>5^{199}\)
\(\Rightarrow\dfrac{1}{3^{300}}< \dfrac{1}{5^{199}}\)
2) a) \(107^{50}=\left(107^2\right)^{25}=11449^{25}\)
\(73^{75}=\left(73^3\right)^{25}=389017^{25}>11449^{25}\)
\(\Rightarrow107^{50}< 73^{75}\)
b) \(54^4< 5^{12}< 21^{12}\Rightarrow54^4< 21^{12}\)
\(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2}\right)^{5.10}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}=\frac{1}{32^{10}}