Question

Answer 1

\(J=\dfrac{2\cdot317\cdot7+14\cdot352+4634}{2+5+8...+65-597}\)

\(=\dfrac{14\cdot317+14\cdot352+14\cdot331}{2+5+8+...+65-597}\)

\(=\dfrac{14\cdot\left(317+352+331\right)}{2+5+8+...+65-597}\)

\(=\dfrac{14\cdot1000}{2+5+8+...+65-597}\)

\(=\dfrac{14000}{2+5+8+...+65-597}\)

Đặt \(A=2+5+8+...+65\)

Số số hạng trong A là: \(\left(65-2\right):3+1=22\)(số)

Tổng A bằng: \(\left(65+2\right)\cdot22:2=737\)

Thay \(A=737\) vào \(J\), ta được:

\(J=\dfrac{14000}{737-597}\)

\(=\dfrac{14000}{140}=100\)

Vậy \(J=100\)

#Ayumu