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\(=\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1\)
\(=\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2008}+\dfrac{2009}{2009}\)
\(=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2009}\right)\)
Bài 2:
a: \(5^{2008}+5^{2007}+5^{2006}\)
\(=5^{2006}\left(5^2+5+1\right)=5^{2006}\cdot31⋮31\)
b: \(8^8+2^{20}\)
\(=2^{24}+2^{20}\)
\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)
Đặt B=\(\frac{2}{4^2}+\frac{2}{6^2}+\frac{2}{8^2}+....+\frac{2}{2008^2}\)
=> A+B= 2\(\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2007^2}+\frac{1}{2008^2}\right)\) <2 \(\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{2006\cdot2007}+\frac{1}{2007\cdot2008}\right)\)
=2\(\left(\frac{1}{2}-\frac{1}{2008}\right)\)=\(\frac{2006}{2008}\)
mà A<B=>A+A<A+B=2006/2008
=>A<1003/2008
mấy câu kia cũng tương tự, mình làm biếng quá
Giải:
1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)
\(=7^8.\left(\dfrac{1}{7}\right)^8\)
\(=7^8.\dfrac{1^8}{7^8}\)
\(=1\)
2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)
\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)
\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)
\(=1\)
3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)
\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)
\(=1\)
4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)
\(=\dfrac{5}{13}\)
Vậy ...
