So Sánh : (1/243)^9 và (1/83)^13
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`(1/243)^9 = [1/(3^5)]^9 = [(1/3)^5]^9=(1/3)^13`
Vì: `1/3 > 1/83`
`=> (1/3)^13 > 1/(83)^13`.
Ta có :
\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)
\(\Rightarrow\frac{1}{243^9}>\frac{1}{8^{13}}\)
Ta có :
\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)
\(\Rightarrow\frac{1}{243^9}>\frac{1}{83^{13}}\)
mình chắc chắn luôn
Sửa đề: \(\left(\dfrac{1}{81}\right)^{13}\)
Ta có: \(\left(\dfrac{1}{243}\right)^9=\left(\dfrac{1}{3}\right)^{45}\)
\(\left(\dfrac{1}{81}\right)^{13}=\left(\dfrac{1}{3}\right)^{52}\)
mà \(\left(\dfrac{1}{3}\right)^{45}< \left(\dfrac{1}{3}\right)^{52}\)
nên \(\left(\dfrac{1}{243}\right)^9< \left(\dfrac{1}{81}\right)^{13}\)
a) \(\left(\frac{1}{243}\right)^9=\left(\frac{1}{3^5}\right)^9=\frac{1}{3^{45}}\)
\(\left(\frac{1}{83}\right)^{13}< \left(\frac{1}{81}\right)^{13}=\left(\frac{1}{3^4}\right)^{13}=\frac{1}{3^{52}}< \frac{1}{3^{45}}=\left(\frac{1}{243}\right)^9\Rightarrow\left(\frac{1}{83}\right)^{13}< \left(\frac{1}{243}\right)^9\)
b) 199010 + 19909
= 19909 ( 1990 + 1 )
= 19909 . 1991 < 199110 = 19919 . 1991
Vậy 199010 + 19909 < 199110
ta co( \(\frac{1}{243}\))9=(\(\frac{1}{3}\))45=(\(\frac{1}{81}\))11,25<(\(\frac{1}{83}\))13
ta co( \(\frac{1}{243}\))9=(\(\frac{1}{3}\))45=(\(\frac{1}{81}\))11,25<(\(\frac{1}{83}\))13
\(\left(\frac{1}{243}\right)^9=\frac{1}{243^9}=\frac{1}{\left(3^5\right)^9}=\frac{1}{3^{45}}\)
\(\left(\frac{1}{83}\right)^{13}< \left(\frac{1}{81}\right)^{13}=\frac{1}{81^{13}}=\frac{1}{\left(3^4\right)^{13}}=\frac{1}{3^{52}}\)
Có \(3^{45}< 3^{52}\Rightarrow\frac{1}{3^{45}}>\frac{1}{3^{52}}\)
suy ra \(\left(\frac{1}{243}\right)^9>\left(\frac{1}{83}\right)^{13}\).