Question

Tìm x biết :

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

Cần cách làm vì mk bt kết quả rồi !

 

Answer 1

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

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)

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

\(\Rightarrow\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}=\frac{10}{30}-\frac{9}{30}=\frac{1}{30}\)

\(\Rightarrow\left(3x+3\right).1=1.30\Rightarrow3x+3=30\Rightarrow3x=27\Rightarrow x=9\)

Answer 2

\(\frac{1}{3}-\frac{1}{3x+3}=\frac{3x+3}{9x+9}-\frac{3}{9x+9}=\frac{3x}{9x+9}=\frac{3}{10}\)

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

\(10x=9x+9\)

\(x=9\)