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\(\left(\frac{1}{243}\right)^9=\left(\frac{1}{3^4}\right)^9=\frac{1}{3^{4.9}}=\frac{1}{3^{36}}\)
\(\left(\frac{1}{83}\right)^{13}<\left(\frac{1}{81}\right)^{13}=\left(\frac{1}{3^4}\right)^{13}=\frac{1}{3^{4.13}}=\frac{1}{3^{42}}\)
\(\frac{1}{3^{36}}>\frac{1}{3^{42}}\Rightarrow\left(\frac{1}{81}\right)^{13}<\left(\frac{1}{243}\right)^9\)
=> \(\left(\frac{1}{83}\right)^{13}<\left(\frac{1}{243}\right)^9\)
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
Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)