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ta có: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(\Rightarrow1-\frac{1}{1+2+3+...+n}=1-1:\frac{n.\left(n+1\right)}{2}=1-\frac{2}{n.\left(n+1\right)}\)
\(=\frac{n.\left(n+1\right)-2}{n.\left(n+1\right)}=\frac{n^2+n-2}{n.\left(n+1\right)}=\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}\) (*)
Từ (*)
\(\Rightarrow1-\frac{1}{1+2}=\frac{4.1}{2.3};1-\frac{1}{1+2+3}=\frac{5.2}{3.4};...;1-\frac{1}{1+2+3+...+n}=\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}\)
\(\Rightarrow E=\frac{4.1}{2.3}.\frac{5.2}{3.4}...\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}=\frac{4.1.5.2...\left(n+1\right).\left(n-2\right).\left(n+2\right).\left(n-1\right)}{2.3.3.4....\left(n-1\right).n.n.\left(n+1\right)}\)\(=\frac{n+2}{n.n}\)
\(\Rightarrow\frac{E}{F}=E:F=\left(\frac{n+2}{n.n}\right):\frac{n+2}{n}=\frac{n+2}{n.n}.\frac{n}{n+2}=\frac{1}{n}\)
\(\Rightarrow\frac{E}{F}=\frac{1}{n}\)
Ta có \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\) (đpcm)
Áp dụng công thức trên ta có
A\(=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdot\cdot\cdot\cdot\cdot\cdot\cdot+\frac{1}{2015\cdot2016\cdot2017}\)
\(\Leftrightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2015\cdot2016\cdot2017}\)
\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{2}{3\cdot4}+....+\frac{1}{2015\cdot2016}-\frac{1}{2016\cdot2017}\)
\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\)
\(\Rightarrow A=\left(\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\right)\div2\approx0.25\)
Vậy A\(\approx0.25\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
a) \(\left|2x-1\right|=5\)
\(\Rightarrow\left[\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\left[\begin{matrix}=3\\=-2\end{matrix}\right.\)
b) \(\left(5^x-1\right)3-2=70\)
\(\Rightarrow5^x.3-3=72\)
\(\Rightarrow5^x.3=75\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
c) \(\left(x-1\frac{1}{2}\right)^2+\frac{3}{4}=\frac{1}{4}\)
\(\Rightarrow\left(x-1\frac{1}{2}\right)^2=\frac{-1}{2}\)
............. Làm tiếp nhé!
d) \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)