\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}+\dfrac{2}{99.101}\)
giúp mình với ạ, giải từng bước nhé ạ
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}\dfrac{2}{7.9}+.........+\dfrac{2}{99.101}\)
\(P=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
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=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
Câu 1:
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
= \(\dfrac{1}{3}-\dfrac{1}{101}\)
= \(\dfrac{98}{303}\)
Câu 2 làm tương tự ở câu 1 nhé
B= \(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) +\(\dfrac{2}{5.7}\) +...+ \(\dfrac{2}{97.99}\) + \(\dfrac{2}{99.101}\)
\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)
Tính nhanh :
\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....+\dfrac{2}{97.99}\)
\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(M=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99})\)
\(M=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(M=2.\dfrac{32}{99}\)
\(M=\dfrac{64}{99}\)
http://vietjack.com/giai-sach-bai-tap-toan-6/bai-95-trang-28-sach-bai-tap-toan-6-tap-2.jsp
\(m=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
a/ \(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
b/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
c/ \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
d/ \(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}>4\)
Chứng minh:
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)> 32 \(^0\)/\(_0\)
Giải:
Biến đổi vế trái BĐT:
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
Vì \(\dfrac{32}{99}>\dfrac{32}{100}\)
\(\Leftrightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}>\dfrac{32}{100}\)
\(\Leftrightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}>32\%\)
Vậy ...
\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
Giúp mình với
Ta có
M = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{99}\)
M = \(\dfrac{32}{99}\)
Vậy M = \(\dfrac{32}{99}\)
\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(M=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(M=\dfrac{1}{3}-\dfrac{1}{99}\)
\(M=\dfrac{32}{99}\)
\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+................+\dfrac{2}{97.99}\)
\(M=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.....................+\dfrac{1}{97}-\dfrac{1}{99}\)
\(M=\dfrac{1}{3}-\dfrac{1}{99}\)
\(M=\dfrac{32}{99}\)
B=\(\dfrac{-1}{3}\)+ \(\dfrac{-1}{3.5}\)+ \(\dfrac{-1}{5.7}\)+ \(\dfrac{-1}{7.9}\) +... + \(\dfrac{-1}{99.101}\)
\(B=-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{-1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{100}{101}=-\dfrac{50}{101}\)
CHO A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+.....+\dfrac{2}{97.99}\)
Tính A
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+.....+\dfrac{2}{97.99}\)
Ta thấy:\(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{1.3
}\)
\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{3.5}\)
............\(\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{2}{97.99}\)
\(\Rightarrow A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..........+\dfrac{1}{97}-\dfrac{1}{99}\) =\(\dfrac{1}{1}-\dfrac{1}{99}\)
=\(\dfrac{99}{99}-\dfrac{1}{99}\)
=\(\dfrac{98}{99}\)
Vậy A=\(\dfrac{98}{99}\)
A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
A = \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(1-\dfrac{1}{99}\)
A = \(\dfrac{98}{99}\)
A=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A=\(1-\dfrac{1}{99}=\dfrac{98}{99}\)
\(\dfrac{6}{2.5}+\dfrac{2}{5.7}+\dfrac{6}{8.11}+...+\dfrac{6}{59.62}+\dfrac{6}{62.65}\)
giúp mình từng bước một nhé
Sửa đề: \(\dfrac{6}{2\cdot5}+\dfrac{6}{5\cdot8}+...+\dfrac{6}{62\cdot65}\)
\(=2\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{62\cdot65}\right)\)
=3(1/2-1/5+1/5-1/8+...+1/62-1/65)
=3*63/130=189/130