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\(\frac{2004}{1}+\frac{2003}{2}+...+\frac{1}{2004}=\left(2004-1-1-...-1\right)+\left(\frac{2003}{2}+1\right)+...+\left(\frac{1}{2004}+1\right)\)
\(=1+\frac{2005}{2}+...+\frac{2005}{2014}=2005\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2004}\right)\)
vậy \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+...+\frac{1}{2004}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{2005\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}\right)}=\frac{1}{2005}\)
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
3. S= -1/6 + -1/20 + 1/10 + 1/6
=0
4. A= -1 -1 -1 -1 -.... -1 [ có (50-2): 2 +1 = 25 số -1)
=-25
=23/25
bạn trình bày ra dc ko Nguyễn Ngọc Bảo Trân