Ta có: \(\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+...+\frac{2}{x^2+3x}=\frac{1}{9}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{2}{3}\cdot\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{2}{3}\cdot\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{2}{3}\cdot\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{x+3}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{4}-\frac{1}{x+3}=\frac{1}{9}:\frac{2}{3}=\frac{1}{9}\cdot\frac{3}{2}=\frac{1}{6}\)
\(\Leftrightarrow-\frac{1}{x+3}=\frac{1}{6}-\frac{1}{4}=\frac{2}{12}-\frac{3}{12}=-\frac{1}{12}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{12}\)
\(\Leftrightarrow x+3=12\)
hay x=9
Vậy: x=9
a có: 114+135+165+...+2x2+3x=19114+135+165+...+2x2+3x=19
⇔23⋅(14−17)+23⋅(17−110)+23⋅(110−113)+...+23⋅(1x−1x+3)=19⇔23⋅(14−17)+23⋅(17−110)+23⋅(110−113)+...+23⋅(1x−1x+3)=19
⇔23⋅(14−17+17−110+110−113+...+1x−1x+3)=19⇔23⋅(14−17+17−110+110−113+...+1x−1x+3)=19
⇔23⋅(14−1x+3)=19⇔23⋅(14−1x+3)=19
⇔14−1x+3=19:23=19⋅32=16⇔14−1x+3=19:23=19⋅32=16
⇔−1x+3=16−14=212−312=−112⇔−1x+3=16−14=212−312=−112
⇔1x+3=112⇔1x+3=112
⇔x+3=12⇔x+3=12
hay x=9
Vậy: x=9