Question

Mọi người giúp em giải bài này với!

Tìm x:\(\frac{3}{2}+\frac{3}{14}+\frac{3}{35}+\frac{3}{65}+........+\frac{6}{\left(x-3\right)x}=\frac{96}{49}\)

Answer 1

 \(\frac{3}{2}+\frac{3}{14}+\frac{3}{15}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)

 \(\frac{6}{\left(1.2\right).2}+\frac{6}{\left(2.7\right).2}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)

 \(\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{\left(x-3\right).x}=\frac{96}{49.2}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{96}{98}\)

=> \(1-\frac{1}{x}=\frac{48}{49}\)

=> \(\frac{1}{x}=\frac{1}{49}\)

=> \(x=49\)