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\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)(Vì 3x + 3 lớn hơn 3x - 1 là 4 đơn vị)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x+1-1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x}{3x+3}=\frac{3}{10}\)
\(\Rightarrow10x=3.\left(3x+3\right)\)
\(\Rightarrow10x=9x+9\)
\(\Rightarrow x=9\)
Vậy...
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-....-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)
Nên 3x + 3 = 30
3x = 30 - 3 = 27
x = 27 : 3 = 9
\(E=\frac{\frac{4}{3\cdot7}-\frac{4}{11.15}}{1-\frac{3}{7}-\frac{3}{11}+\frac{1}{5}}-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}\right)\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{11}+\frac{1}{15}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2006}-\frac{1}{2007}\right)\)
\(=\frac{\frac{64}{385}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{2007}\right)\)
\(=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{2007}\right)=\frac{1}{2007}\)
Vậy : \(E=\frac{1}{2007}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}=\frac{10}{30}-\frac{9}{30}=\frac{1}{30}\)
\(\Rightarrow\left(3x+3\right).1=1.30\Rightarrow3x+3=30\Rightarrow3x=27\Rightarrow x=9\)
Thôi được rồi .
Giải:
\(P=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{\left(4n-3\right)\left(4n+1\right)}\)
\(\Rightarrow4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{\left(4n-3\right)\left(4n+1\right)}\)
\(=\frac{5-1}{1.5}+\frac{9-5}{5.9}+...+\frac{\left(4n+1\right)-\left(4n-3\right)}{\left(4n-3\right)\left(4n+1\right)}\)
\(=\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+...+\frac{1}{4n-3}-\frac{1}{4n+1}\)
\(=1-\frac{1}{4n+1}=\frac{4n}{4n+1}\)
Vậy \(A=\frac{4n}{4n+1}\)
\(A=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{81\cdot85}+\frac{5}{85\cdot89}\\ A=\frac{5}{4}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{81\cdot85}+\frac{4}{85\cdot89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{81}-\frac{1}{85}+\frac{1}{85}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{89}{267}-\frac{3}{267}\right)\\ A=\frac{5}{4}\cdot\frac{86}{267}=\frac{215}{534}\)