\(a)\dfrac{11}{5.7}+\dfrac{11}{7.9}+\dfrac{11}{9.11}+...+\dfrac{11}{59.61} \)
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14 tháng 3 2017
\(T=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{59\cdot61}\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{59\cdot61}\right)\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)
14 tháng 3 2017
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{59.61}\)
\(=3.\left(\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{59.61}\right)\)
\(=3.\dfrac{1}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}.\dfrac{56}{305}\)
\(=\dfrac{84}{305}\)
MV
12 tháng 8 2017
a,
\(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\\ =1\cdot\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\\ =\left(2-1\right)\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\\ =\left(2-1\right)\cdot\dfrac{1}{2^2}+\left(2-1\right)\cdot\dfrac{1}{2^3}+...+\left(2-1\right)\cdot\dfrac{1}{2^{2006}}\\ =\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{2005}}-\dfrac{1}{2^{2006}}\\ =\dfrac{1}{2}-\dfrac{1}{2^{2006}}\\ =\dfrac{2^{2005}}{2^{2006}}-\dfrac{1}{2^{2006}}\\ =\dfrac{2^{2005}-1}{2^{2006}}\)
b,
\(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\\ =\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\\ =\dfrac{1}{5}-\dfrac{1}{61}\\ =\dfrac{56}{305}\)
c,
\(\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\\ =\dfrac{7}{2}\cdot\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{9999}\right)\\ =\dfrac{7}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{7}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{7}{2}\cdot\left(1-\dfrac{1}{101}\right)\\ =\dfrac{7}{2}\cdot\dfrac{100}{101}\\ =\dfrac{350}{101}\)
12 tháng 8 2017
Đặt:
\(X=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
\(2X=2\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)
\(2X=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(2X-X=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)\(X=\dfrac{1}{2}-\dfrac{1}{2^{2016}}\)
\(Y=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{59.61}\)
\(Y=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(Y=\dfrac{1}{5}-\dfrac{1}{61}=\dfrac{56}{305}\)
\(Z=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)
\(Z=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{99.101}\)
\(Z=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(Z=\dfrac{7}{2}\left(1-\dfrac{1}{101}\right)\)
\(Z=\dfrac{7}{2}.\dfrac{100}{101}=\dfrac{700}{202}\)
OY
22 tháng 7 2021
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
22 tháng 7 2021
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
25 tháng 4 2018
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
=\(\dfrac{3}{2}.\dfrac{56}{305}\)
= \(\dfrac{78}{305}\)
25 tháng 4 2018
\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)
*Nếu \(x^2-4=0\)
⇒ x2 = 4
⇒ x ∈ {2 ; -2}
*Nếu \(6-2x=0\)
⇒2x = 6
⇒ x = 6 : 2 = 3
Vậy x ∈ { -2 ; 2 ; 3 }
PT
13 tháng 8 2017
a) Sửa tí: \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
\(\Rightarrow2A=2.\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(\Rightarrow2A-A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2006}}\right)\)
\(\Rightarrow A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2006}}\)
\(\Rightarrow A=2-\dfrac{1}{2^{2006}}\)
b) Đặt \(A=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{1}{50.61}\)
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(A=\dfrac{1}{5}-\dfrac{1}{61}\)
\(A=\dfrac{56}{305}\)
c) Đặt \(A=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)
\(A=\dfrac{7}{2}.2.\left(\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{9999}\right)\)
\(A=\dfrac{7}{2}.\left(1-\dfrac{1}{101}\right)\)
\(A=\dfrac{7}{2}.\dfrac{100}{101}\)
\(A=\dfrac{256}{101}\)
27 tháng 3 2017
Theo quy luật thì mình nghĩ đáng lẽ \(\dfrac{4}{5.9}\)phải là\(\dfrac{4}{7.9}\)Bạn có chép sai đề ko?
27 tháng 3 2017
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=\(\dfrac{112}{305}\) \(\Rightarrow A=1-\dfrac{112}{305}=\dfrac{193}{305}\)
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`