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\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Gọi A = \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
=> A = \(\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
A < \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A < \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
A < \(\frac{1}{2}-\frac{1}{100}\)
A < \(\frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
=> A < \(\frac{1}{2}\)
<=> \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
\(3^{2^{3^2}}=9^6\)
\(2^{3^{2^3}}=8^6\)
Vì \(9^6>8^6\)
\(\Rightarrow3^{2^{3^2}}>2^{3^{2^3}}\)
3^2^3^2<2^3^2^3
chắc zậy mà mink cũng ko chắc đâu nha!!!
\(a,2^n\cdot4=128\\ \Rightarrow2^n=32\\ \Rightarrow n=5\\ b,\Rightarrow\left(2^n+1\right)^3=5^3\\ \Rightarrow2^n+1=5\\ \Rightarrow2^n=4\Rightarrow n=2\\ c,n^{15}=n\\ \Rightarrow n^{15}-n=0\\ \Rightarrow n\left(n^{14}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}n=0\\n^{14}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=0\\n=1\\n=-1\end{matrix}\right.\)
20191 = 2019
hok tốt
20191 = 2019
~Study well~
#SJ