Question

\(\text{A=}\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)

Loi giai day du gium minh nha may ban

minh dang gap

 

Answer 1

\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+.......+\frac{3}{97.100}\right)\)

\(=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{97}-\frac{1}{100}\right)\)

\(=3.\left(1-\frac{1}{100}\right)\)

\(=3.\frac{99}{100}\)

\(=\frac{297}{100}\)

Answer 2

Dễ thôi bạn mẫu cách nhau 3 đơn vị tử xuất hiện 3 chỉ cần rút rọn đi 3 là tử có nhé

Ta có: \(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+....+\frac{3^2}{97.100}\)

\(\frac{1}{3}A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.......+\frac{3}{97.100}\)

\(\frac{1}{3}A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)

\(\frac{1}{3}A=1-\frac{1}{100}\)

\(\frac{1}{3}A=\frac{99}{100}\)

\(A=\frac{99}{100}.3=\frac{297}{100}\)