Question

Cho \(A=\frac{1}{3x63}+\frac{1}{4x64}+\frac{1}{5x65}+....+\frac{1}{1802x1862}\)

       \(B=\frac{1}{3x1803}+\frac{1}{4x1804}+.....+\frac{1}{62x1862}\)

Tính \(\frac{A}{B}\)

Answer 1

minh giải giống Đinh Thùy Linh nhưng câu cuối (câu kết luận) phải là:

Vậy \(\frac{60.A}{1800.B}=1\Rightarrow\frac{A}{B}=\frac{1800}{60}=30\)

Answer 2

\(60A=\frac{1}{3}-\frac{1}{63}+\frac{1}{4}-\frac{1}{64}+...+\frac{1}{1802}-\frac{1}{1862}=\)

\(=\frac{1}{3}+\frac{1}{4}+...+\frac{1}{62}-\left(\frac{1}{1803}+\frac{1}{1804}+...+\frac{1}{1862}\right).\)

\(1800B=\frac{1}{3}-\frac{1}{1803}+\frac{1}{4}-\frac{1}{1804}+...+\frac{1}{62}-\frac{1}{1862}\)\(=\frac{1}{3}+\frac{1}{4}+...+\frac{1}{62}-\left(\frac{1}{1803}+\frac{1}{1804}+...+\frac{1}{1862}\right).\)

Vậy: \(\frac{30A}{1800B}=1\Rightarrow\frac{A}{B}=60\).