a) Ta có: \(x-\dfrac{2}{3\cdot5}-\dfrac{2}{5\cdot7}-\dfrac{2}{7\cdot9}-\dfrac{2}{9\cdot11}-\dfrac{2}{11\cdot13}-\dfrac{2}{13\cdot15}=\dfrac{2}{5}\)
\(\Leftrightarrow x\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow x\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow x\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow x\cdot\dfrac{4}{15}=\dfrac{2}{5}\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
b) Ta có: \(\dfrac{1}{2\cdot3}\cdot x+\dfrac{1}{3\cdot4}\cdot x+\dfrac{1}{4\cdot5}\cdot x+...+\dfrac{1}{49\cdot50}\cdot x=1\)
\(\Leftrightarrow x\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{49\cdot50}\right)=1\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)
\(\Leftrightarrow x\cdot\dfrac{12}{25}=1\)
hay \(x=\dfrac{25}{12}\)
Vậy: \(x=\dfrac{25}{12}\)
