Question

Tìm x :

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)\) . \(x\) = \(\frac{22}{45}\)

Thanks truoc

Answer 1

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)

\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+......+\frac{2}{8.9.10}\right).x=\frac{22}{45}\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{22}{45}\)

\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{22}{45}\)

\(\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)

\(\frac{11}{45}.x=\frac{22}{45}\)

\(x=\frac{22}{45}:\frac{11}{45}\)

\(x=2\)