Question

(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+....\(\dfrac{1}{8.9.10}\)) . x = \(\dfrac{22}{45}\)

Tìm x nha mọi người mình đang cần gấp lắm ạ

Answer 1

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

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

=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)

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

=> \(\dfrac{1}{2}.x=1\)

=> \(x=2\)

Vậy x = 2

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