TA CÓ THỂ THẤY, VẾ TRÁI CÓ: 12 CẶP
=> \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
<=> \(x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\) (****)
Ta xét: \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
=> \(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\)
=> \(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\)
=> \(2A=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Ta tiếp tục xét: \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
=> \(3B=1+\frac{1}{3}+...+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)
=> \(2B=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(B=\frac{121}{243}\)
THAY CÁC GIÁ TRỊ A; B VÀO PT (****) TA ĐƯỢC:
=> \(x+\frac{12}{25}=\frac{121}{243}\)
<=> \(x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
