Question

Answer 1

\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{618}\)

\(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{x.\left(x+3\right)}=\dfrac{101}{206}\)

\(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{101}{206}\)

\(\dfrac{1}{2}-\dfrac{1}{x+3}=\dfrac{101}{206}\)

\(\dfrac{1}{x+3}=\dfrac{1}{103}\)

x+3=103

x=100

Answer 2

Answer 3

Ta có: \(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{618}\)

\(\Leftrightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{101}{206}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+3}=\dfrac{101}{206}\)

\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{103}\)

\(\Leftrightarrow x+3=103\)

hay x=100