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\(\left(\frac{1}{243}\right)^9=\left(\frac{1}{3^4}\right)^9=\frac{1}{3^{4.9}}=\frac{1}{3^{36}}\)
\(\left(\frac{1}{83}\right)^{13}<\left(\frac{1}{81}\right)^{13}=\left(\frac{1}{3^4}\right)^{13}=\frac{1}{3^{4.13}}=\frac{1}{3^{42}}\)
\(\frac{1}{3^{36}}>\frac{1}{3^{42}}\Rightarrow\left(\frac{1}{81}\right)^{13}<\left(\frac{1}{243}\right)^9\)
=> \(\left(\frac{1}{83}\right)^{13}<\left(\frac{1}{243}\right)^9\)
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
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
a) 80<243 nên \(\frac{1}{80}>\frac{1}{243}\)
\(\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^7\) mà \(\left(\frac{1}{243}\right)^7>\left(\frac{1}{243}\right)^6\)
Nên \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
b) Ta so sánh \(\frac{3}{8}\) với \(\frac{5}{243}\)
Ta có: \(\frac{3}{8}=\frac{3.243}{8.243}=\frac{729}{1944}\)
\(\frac{5}{243}=\frac{5.8}{243.8}=\frac{40}{1944}\)
Suy ra: \(\frac{3}{8}>\frac{5}{243}\)
\(\Rightarrow\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^5\) mà \(\left(\frac{5}{243}\right)^5>\left(\frac{5}{243}\right)^3\)
Nên \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)