Question

\(\frac{3}{2x5}\)+ \(\frac{3}{5x8}\)+ \(\frac{3}{8x11}\)+ ... + \(\frac{3}{\left(X-3\right)xX}\)= \(\frac{1}{6}\)

Giúp mk nhé!!!

Answer 1

Giải:

Ta có:

\(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{\left(x-3\right)\times x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x-3}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\Leftrightarrow x=3\)

Answer 2

De thoi ma,minh ko ghi lai de nha

=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+...+...=\(\frac{1}{6}\)

=\(\frac{1}{2}\)-\(\frac{1}{x}\)=\(\frac{1}{6}\)

Con lai bn tu lam nha . chuc bn hok tot !!!

Answer 3

\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+....+\frac{3}{\left(x-3\right).x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{3}\)

\(\Leftrightarrow x=1\div\frac{1}{3}=3\)