Question

[super saiyangoku]

Tìm x biết 

a  1/5×8+1/8×11+1/11×14 ....+1/x×(x+3)=101/1540

b  1+1/3+1/6+1/10+ ....+1/x×(x+1)÷2=1/1991/1993

Answer 1

[Thắng Nguyễn]

a)\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)

\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)

b ko hiểu đề