Tính \(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+............+\dfrac{4}{59.61}\)
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Tính giá trị các biểu thức:
\(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
Đặt A=\(\frac{4}{5.7}\)+\(\frac{4}{7.9}\)+...+\(\frac{4}{59.61}\)
A=2( \(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+...+\(\frac{2}{59.61}\))
A=2( \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\)\(\frac{1}{59}-\frac{1}{61}\))
=2( \(\frac{1}{5}-\frac{1}{61}\))=2.\(\frac{56}{305}\)=\(\frac{112}{305}\)
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So sánh:
A = \(\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+\dfrac{3^2}{8.11}\) và B = \(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
So sánh: A =\(\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+\dfrac{3^2}{8.11}\) và B \(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
Tính giá trị các biểu thức:
\(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
Ta có:
\(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
\(\dfrac{A}{2}=\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
\(\dfrac{A}{2}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(\dfrac{A}{2}=\dfrac{1}{6}-\dfrac{1}{61}\)
\(\dfrac{A}{2}=\dfrac{56}{305}\)
\(\Rightarrow A=\dfrac{112}{305}\)
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\(a)\dfrac{11}{5.7}+\dfrac{11}{7.9}+\dfrac{11}{9.11}+...+\dfrac{11}{59.61} \)
`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`
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`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`
`= 616/305 : 2`
`= 308/305`
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Tính một cách hợp lí :
a) \(4\dfrac{3}{4}+\left(-0.37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
b) \(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
c) \(\dfrac{\dfrac{5}{22}+\dfrac{3}{13}-\dfrac{1}{2}}{\dfrac{4}{13}-\dfrac{2}{11}+\dfrac{3}{2}}\)
Tính nhanh tổng sau
A= \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
\(A=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{61-59}{59.61}\)
\(A=\dfrac{5}{3.5}-\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7}+\dfrac{9}{7.9}-\dfrac{7}{7.9}+...+\dfrac{61}{59.61}-\dfrac{59}{59.61}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(A=\dfrac{1}{3}-\dfrac{1}{61}=\dfrac{61}{183}-\dfrac{3}{183}=\dfrac{58}{183}\)
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\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
= \(\dfrac{1}{3}-\dfrac{1}{61}\)
= \(\dfrac{58}{183}\)
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A= \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
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= \(\dfrac{1}{3}-\dfrac{1}{61}\)
A= \(\dfrac{58}{183}\)
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Xem thêm câu trả lời
tính một cách hợp lí:
A) \(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
b) \(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
c) \(\dfrac{\dfrac{5}{22}+\dfrac{3}{13}-\dfrac{1}{2}}{\dfrac{4}{13}-\dfrac{2}{11}+\dfrac{3}{2}}\)
b,=1/5-1/7+1/7-1/9+...+1/59-1/61
=1/5-1/61
=54/115
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Tính :
A = 1 - \(\dfrac{4}{5.7}-\dfrac{4}{5.9}-\dfrac{4}{9.11}-......-\dfrac{4}{59.61}\)
Theo quy luật thì mình nghĩ đáng lẽ \(\dfrac{4}{5.9}\)phải là\(\dfrac{4}{7.9}\)Bạn có chép sai đề ko?
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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=\(\dfrac{112}{305}\) \(\Rightarrow A=1-\dfrac{112}{305}=\dfrac{193}{305}\)
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