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Bước 1: Gom nhóm và chuyển đổi mẫu số
Ta viết lại biểu thức như sau:
A = 7 / (31 × 19) + 5 / (43 × 19) + 3 / (43 × 23) + 11 / (57 × 23) + 19 / (57 × 38)
Bước 2: Biến đổi thành tổng các sai phân
A = 2 × [ 7 / (31 × 38) + 5 / (38 × 43) + 3 / (43 × 46) + 11 / (46 × 57) + 19 / (57 × 76) ]
Bước 3: Áp dụng công thức sai phân
• 7 / (31 × 38) = (1/31 - 1/38)
• 5 / (38 × 43) = (1/38 - 1/43)
• 3 / (43 × 46) = (1/43 - 1/46)
• 11 / (46 × 57) = (1/46 - 1/57)
• 19 / (57 × 76) = (1/57 - 1/76)
Cộng lại:
A = 2 × [ ( (1/31 - 1/38)) + ( (1/38 - 1/43)) + ( (1/43 - 1/46)) + ( (1/46 - 1/57)) + (((1/57 - 1/76)) ]
Rút gọn biểu thức:
A = 2 × (1/31 - 1/76) = 2 × (45 / (31 × 76)) = 90 / 2356 = 45 / 1178
Kết luận:
A = 45 / 1178
Ta có: A=47.31 +67.41 +910.41 +710.57
=2035.31 +3035.41 +4550.41 +3550.57
=5(431.35 +635.41 +941.50 +750.57 )
=5(131 −135 +135 −141 +141 −150 +150 −157 )
=5(131 −157 )
B=719.31 +519.43 +323.43 +1123.57
=1438.31 +1038.43 +646.43 +2246.57
=2(731.38 +538.43 +343.46 +1146.57 )
=2(131 −138 +138 −143 +143 −146 +146 −157 )
=2(131 −157 )
⇒AB =5(131 −157 )2(131 −157 ) =52
Ta có: \(A=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)
\(=\frac{20}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}\)
\(=5\left(\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\right)\)
\(=5\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)
\(=5\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(B=\frac{7}{19.31}+\frac{5}{19.43}+\frac{3}{23.43}+\frac{11}{23.57}\)
\(=\frac{14}{38.31}+\frac{10}{38.43}+\frac{6}{46.43}+\frac{22}{46.57}\)
\(=2\left(\frac{7}{31.38}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{5\left(\frac{1}{31}-\frac{1}{57}\right)}{2\left(\frac{1}{31}-\frac{1}{57}\right)}=\frac{5}{2}\)
Ta có: \(A = \frac{4}{7.31} + \frac{6}{7.41} + \frac{9}{10.41} + \frac{7}{10.57}\)
\(= \frac{20}{35.31} + \frac{30}{35.41} + \frac{45}{50.41} + \frac{35}{50.57}\)
\(= 5 \left(\right. \frac{4}{31.35} + \frac{6}{35.41} + \frac{9}{41.50} + \frac{7}{50.57} \left.\right)\)
\(= 5 \left(\right. \frac{1}{31} - \frac{1}{35} + \frac{1}{35} - \frac{1}{41} + \frac{1}{41} - \frac{1}{50} + \frac{1}{50} - \frac{1}{57} \left.\right)\)
\(= 5 \left(\right. \frac{1}{31} - \frac{1}{57} \left.\right)\)
\(B = \frac{7}{19.31} + \frac{5}{19.43} + \frac{3}{23.43} + \frac{11}{23.57}\)
\(= \frac{14}{38.31} + \frac{10}{38.43} + \frac{6}{46.43} + \frac{22}{46.57}\)
\(= 2 \left(\right. \frac{7}{31.38} + \frac{5}{38.43} + \frac{3}{43.46} + \frac{11}{46.57} \left.\right)\)
\(= 2 \left(\right. \frac{1}{31} - \frac{1}{38} + \frac{1}{38} - \frac{1}{43} + \frac{1}{43} - \frac{1}{46} + \frac{1}{46} - \frac{1}{57} \left.\right)\)
\(= 2 \left(\right. \frac{1}{31} - \frac{1}{57} \left.\right)\)
\(\Rightarrow \frac{A}{B} = \frac{5 \left(\right. \frac{1}{31} - \frac{1}{57} \left.\right)}{2 \left(\right. \frac{1}{31} - \frac{1}{57} \left.\right)} = \frac{5}{2}\)