Question

BT2: Tìm x\(\in\) N*, biết

2) \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{x.\left(x+1\right)}=\dfrac{299}{600}\)

Answer 1

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{299}{600}\)

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{299}{600}\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{300}{600}-\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{1}{600}\)

=> x + 1 = 600

x = 600 - 1

x = 599

Vậy x = 599