\(\frac{121212}{161616}-\left(\frac{151515}{323232}-x\right)=2\)
=> \(\frac{3}{4}-\left(\frac{15}{32}-x\right)=2\)
=> \(\frac{15}{32}-x=\frac{3}{4}-2\)
=> \(\frac{15}{32}-x=-\frac{5}{4}\)
=> \(x=\frac{15}{32}-\frac{-5}{4}=\frac{15}{32}+\frac{5}{4}=\frac{55}{32}\)
b) \(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+\frac{x}{30}+\frac{x}{42}+\frac{x}{56}+\frac{x}{72}+\frac{x}{90}=\frac{9}{5}\)
=> \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}+\frac{x}{6\cdot7}+\frac{x}{7\cdot8}+\frac{x}{8\cdot9}+\frac{x}{9\cdot10}=\frac{9}{5}\)
=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{9}-\frac{x}{10}=\frac{9}{5}\)
=> \(\frac{x}{1}-\frac{x}{10}=\frac{9}{5}\)
=> \(\frac{10x-x}{10}=\frac{9}{5}\)
=> \(\frac{9x}{10}=\frac{9}{5}\)
=> \(\frac{9x}{10}=\frac{9\cdot2}{5\cdot2}=\frac{18}{10}\)
=> x = 2
