\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x(x+3)}=\dfrac{6}{19}\)
\(\Rightarrow\)\(\dfrac{1}{3}.(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{x(x+3)})=\dfrac{6}{19}\)
\(\Rightarrow\)\(\dfrac{1}{3}.(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3})=\dfrac{6}{19}\)
\(\Rightarrow\)\(\dfrac{1}{3}.(\dfrac{1}{1}-\dfrac{1}{x+3})=\dfrac{6}{19}\)
\(\Rightarrow\) \(\dfrac{1}{1}-\dfrac{1}{x+3}=\dfrac{6}{19}:\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{1}{1}-\dfrac{1}{x+3}=\dfrac{18}{19}\)
\(\Rightarrow\) \(\dfrac{1}{x+3}=\dfrac{1}{1}-\dfrac{18}{19}\)
\(\Rightarrow\) \(\dfrac{1}{x+3}=\dfrac{1}{19}\)
\(\Rightarrow\) \(x+3=19\)
\(x=19-3\)
\(x=16\)
Vậy \(x=16\)
Ta chi can tach ra , xong ta luoc bot,neu co so bi thua ra thi ta tinh tong.....
Ta có:
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{x\left(x+3\right)}=\dfrac{6}{19}\)
\(\Rightarrow\) \(\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+......+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{6}{19}\)
\(\Rightarrow\) \(\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{6}{19}\)
\(\Rightarrow\) \(\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{6}{19}\)
\(\Rightarrow\) \(1-\dfrac{1}{x+3}=\dfrac{6}{19}:\dfrac{1}{3}\)
\(\Rightarrow\) \(1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
\(\Rightarrow\) \(\dfrac{1}{x+3}=1-\dfrac{18}{19}=\dfrac{1}{18}\)
\(\Rightarrow\) x + 3 = 18
\(\Rightarrow\) x = 18 - 3 = 15
Vậy x = 15
\(\dfrac{1}{3}\)(1-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{10}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{\left(x+3\right)}\))=\(\dfrac{6}{19}\)
=\(\dfrac{1}{3}\)(1-\(\dfrac{1}{\left(x+3\right)}\))=\(\dfrac{6}{19}\)
1-\(\dfrac{1}{\left(x+3\right)}\)=\(\dfrac{6}{19}\):\(\dfrac{1}{3}\)=\(\dfrac{18}{19}\)
\(\dfrac{1}{\left(x+3\right)}\)=1-\(\dfrac{18}{19}\)=\(\dfrac{1}{19}\)
suy ra x+3=19
x=19-3
x=16
