Question

Tìm x biết : 

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

Answer 1

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)(Vì 3x + 3 lớn hơn 3x - 1 là 4 đơn vị)

\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{x+1-1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{x}{3x+3}=\frac{3}{10}\)

\(\Rightarrow10x=3.\left(3x+3\right)\)

\(\Rightarrow10x=9x+9\)

\(\Rightarrow x=9\)

Vậy...

Answer 2

thanks