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Em nhớ nhân 1/2 trong tất cả dấu bằng thì biểu thức này mới không thay đổi kết quả nhé.
`11/(5.7) + 11/(7.9) + 11/(9.11) + ... + 11/(59.61)`
`= 2.(11/(5.7) + 11/(7.9) + ... + 11/(59.61))`
`= 11.(2/(5.7) + 2/(7.9) + ... + 2/(59.61))`
`= 11.(1/5 - 1/7 + 1/7 - 1/9 + ... +1/59 - 1/61)`
`= 11.(1/5 - 1/61)`
`= 11.56/305`
`= 616/305`
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
\(A=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{61-59}{59.61}\)
\(A=\dfrac{5}{3.5}-\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7}+\dfrac{9}{7.9}-\dfrac{7}{7.9}+...+\dfrac{61}{59.61}-\dfrac{59}{59.61}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(A=\dfrac{1}{3}-\dfrac{1}{61}=\dfrac{61}{183}-\dfrac{3}{183}=\dfrac{58}{183}\)
Theo quy luật thì mình nghĩ đáng lẽ \(\dfrac{4}{5.9}\)phải là\(\dfrac{4}{7.9}\)Bạn có chép sai đề ko?
A=1-\(\dfrac{4}{5.7}-\dfrac{4}{7.9}-\dfrac{4}{9.11}...-\dfrac{4}{59.61}\)
A=\(1-\left(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\right)\)
Đặt B=\(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\)
B=\(2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{59.61}\right)\) B=\(2\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\) B=\(2\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=2.\dfrac{56}{305}\) B=\(\dfrac{112}{305}\) \(\Rightarrow A=1-\dfrac{112}{305}=\dfrac{193}{305}\)
\(T=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{59\cdot61}\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{59\cdot61}\right)\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{59.61}\)
\(=3.\left(\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{59.61}\right)\)
\(=3.\dfrac{1}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}.\dfrac{56}{305}\)
\(=\dfrac{84}{305}\)
A=3/4.(1/5.7+1/7.9+....+1/59.61)
A=3/4.(1/5-1/7+1/7-1/9+...+1/59-1/61)
A=3/4.(1/5-1/61)
A=3/4.56/305
A=42/305
mình làm cho bạn phần A thôi nhé còn phần B mình chưa nghĩ ra cách làm ahihi!
Ta có :
\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+............+\dfrac{4}{59.61}\)
\(\dfrac{A}{2}=\dfrac{2}{5.7}+\dfrac{2}{7.9}+..............+\dfrac{2}{59.61}\)
\(\dfrac{A}{2}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.......+\dfrac{1}{59}-\dfrac{1}{61}\)
\(\dfrac{A}{2}=\dfrac{1}{5}-\dfrac{1}{61}\)
\(\dfrac{A}{2}=\dfrac{56}{305}\)
\(\Rightarrow A=\dfrac{112}{305}\)
Chúc bn học tốt!!
\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
\(A=2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
\(A=2\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(A=2\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(A=2.\dfrac{56}{305}\)
\(A=\dfrac{112}{305}\)