\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
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Tính \(Q=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...............+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+.............+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
Ta rút gọn phần mẫu:
\(\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}\\ =\left(\dfrac{1}{1}+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{99}\right)+\left(\dfrac{1}{5}+\dfrac{1}{99}\right)+...+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+\left(\dfrac{1}{1}+\dfrac{1}{99}\right)\\ =\dfrac{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{100}\\ =\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{50}\)
Vậy:
\(Q=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\\ =\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{50}}\\ =50\)
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@Nguyễn Huy Tú, @Hoàng Thị Ngọc Anh, @Tuấn Anh Phan Nguyễn, @Hoang Hung Quan, @ngonhuminh, và các bn khác giúp mk với!!
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tính Q = \(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(Q=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2}{1.99}+\dfrac{2}{3.97}+...+\dfrac{2}{51.49}}\)
\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...+\dfrac{100}{51.49}}\)
\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{\dfrac{1+99}{1.99}+\dfrac{3+97}{3.97}+...+\dfrac{51+49}{51.49}}\)
\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{\dfrac{1}{99}+1+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{51}+\dfrac{1}{49}}\)
\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{1+\dfrac{1}{3}+...+\dfrac{1}{99}}\)
\(\Rightarrow Q=50\)
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Cho \(A=1+\dfrac{1}{3}+\dfrac{1}{5}+.......+\dfrac{1}{99}\)
\(B=\dfrac{1}{1.99}+\dfrac{1}{3.97}+.......+\dfrac{1}{97.3}+\dfrac{1}{99.1}\)
Tính A : B
Ta có: \(A=1+\dfrac{1}{3}+\dfrac{1}{5}+.....+\dfrac{1}{99}\)
\(B=\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+......+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)\)
\(=\left(\dfrac{99}{99}+\dfrac{1}{99}\right)+\left(\dfrac{97}{3.97}+\dfrac{3}{3.97}\right)+.....+\left(\dfrac{51}{49.51}+\dfrac{49}{49.51}\right)\)
\(=\dfrac{100}{1.99}+\dfrac{100}{3.97}+......+\dfrac{100}{49.51}\)
\(=100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.....+\dfrac{1}{49.51}\right)\) (1)
Ta có: \(B=\dfrac{1}{1.99}+\dfrac{1}{3.97}+......+\dfrac{1}{97.3}+\dfrac{1}{99.1}\)
\(=\left(\dfrac{1}{1.99}+\dfrac{1}{99.1}\right)+\left(\dfrac{1}{3.97}+\dfrac{1}{97.3}\right)+......+\left(\dfrac{1}{49.51}+\dfrac{1}{51.49}\right)\)
\(=\dfrac{2}{1.99}+\dfrac{2}{3.97}+......+\dfrac{2}{49.51}\)
\(=2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+......+\dfrac{1}{49.51}\right)\) (2)
Từ (1) và (2) => \(A:B=\dfrac{100}{2}=50\)
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Tính giá trị của biểu thức:
\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.99}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{99}+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+....+\dfrac{1}{99}}\)
Ta có:
\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\dfrac{99+1}{1\cdot99}+\dfrac{97+3}{3\cdot97}+...+\dfrac{1+99}{99\cdot1}}{100}}\)
\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{99}+1\right)}{100}}\)
\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{100}}=\dfrac{1}{\dfrac{2}{100}}=\dfrac{100}{2}=50\)
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\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+...+\dfrac{1}{99}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{1+\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{100}{100}+\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)
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Bài 1: Tính tổng 100 số hạng đầu tiên của các dãy sau:
a)dfrac{1}{2};dfrac{1}{6};dfrac{1}{12};dfrac{1}{20};dfrac{1}{30};...
b)dfrac{1}{6};dfrac{1}{66};dfrac{1}{176};dfrac{1}{336};...
Bài 2: Tính:
a)Adfrac{1+dfrac{1}{3}+dfrac{1}{5}+...+dfrac{1}{97}+dfrac{1}{99}}{dfrac{1}{1.99}+dfrac{1}{3.97}+dfrac{1}{5.95}+...+dfrac{1}{97.3}+dfrac{1}{99.1}}
b)Bdfrac{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{100}}{dfrac{99}{1}+dfrac{98}{2}+dfrac{97}{3}+...+dfrac{1}{99}}
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Bài 1: Tính tổng 100 số hạng đầu tiên của các dãy sau:
a)\(\dfrac{1}{2};\dfrac{1}{6};\dfrac{1}{12};\dfrac{1}{20};\dfrac{1}{30};...\)
b)\(\dfrac{1}{6};\dfrac{1}{66};\dfrac{1}{176};\dfrac{1}{336};...\)
Bài 2: Tính:
a)A=\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
b)B=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}}\)
Bài 1: Tính tổng 100 số hạng đầu tiên của các dãy sau:
a) \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{1}{1.2}\\\dfrac{1}{6}=\dfrac{1}{2.3}\\\dfrac{1}{12}=\dfrac{1}{3.4}\\...\end{matrix}\right.\)
Vậy số thứ 100 của dãy là: \(\dfrac{1}{100.101}=\dfrac{1}{10100}\)
Tổng: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
b) \(\left\{{}\begin{matrix}\dfrac{1}{6}=\dfrac{1}{\left(5.0+1\right)\left(5.1+1\right)}\\\dfrac{1}{66}=\dfrac{1}{\left(5.1+1\right)\left(5.2+1\right)}\\\dfrac{1}{176}=\dfrac{1}{\left(5.2+1\right)\left(5.3+1\right)}\\...\end{matrix}\right.\)
Vậy số thứ 100 của dãy là: \(\dfrac{1}{\left(5.99+1\right)\left(5.100+1\right)}=\dfrac{1}{248496}\)
Tổng: \(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{496.501}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{496.501}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}.\dfrac{500}{501}\)
\(=\dfrac{100}{501}\)
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Bài 2: Tính:
a) \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(A=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(A=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(A=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(\Rightarrow A=\dfrac{100}{2}=50\)
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Bài 2 :
a, Xét tử số : Đặt B = \(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}\)
Số số hạng của tử số là : ( 99 - 1 ) : 2 + 1 = 50 ( số )
=> Tử số có 50 phân số
Ta có : \(B=\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+\left(\dfrac{1}{5}+\dfrac{1}{95}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)\)
\(=\left(\dfrac{99}{99}+\dfrac{1}{99}\right)+\left(\dfrac{97}{3.97}+\dfrac{3}{3.97}\right)+\left(\dfrac{95}{5.95}+\dfrac{5}{5.95}\right)+...+\left(\dfrac{51}{49.51}+\dfrac{49}{49.51}\right)\)
\(=\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{49.51}\)
Xét mẫu số : Đặt C = \(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}\)
\(=\left(\dfrac{1}{1.99}+\dfrac{1}{99.1}\right)+\left(\dfrac{1}{3.97}+\dfrac{1}{97.3}\right)+...+\left(\dfrac{1}{49.51}+\dfrac{1}{51.49}\right)\)
\(=2.\dfrac{1}{1.99}+2.\dfrac{1}{3.97}+...+2.\dfrac{1}{49.51}\)
\(=2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)\)
Thay B và C vào A ta có :
\(A=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(\Rightarrow A=\dfrac{100}{2}=50\)
Vậy A = 50
b, Xét mẫu số : Đặt C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
\(=\dfrac{100-1}{1}+\dfrac{100-2}{2}+\dfrac{100-3}{3}+...+\dfrac{100-99}{99}\)
\(=100-1+\dfrac{100}{2}-1+\dfrac{100}{3}-1+...+\dfrac{100}{99}-1\)
\(=\left(100+\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}\right)-\left(1+1+...+1\right)\)
Đặt D = 1 + 1 + ... + 1
Số số hạng của tổng D là : ( 99 - 1 ) : 1 + 1 = 99 ( số hạng )
\(\Rightarrow D=1.99=99\)
Thay D = 99 ta có :
\(C=100\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-99\)
\(=100+100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-99\)
\(=\left(100-99\right)+100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)\)
\(=1+100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)\)
\(=\dfrac{100}{100}+100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)=100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)
Thay vào đề bài , ta có :
\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)
Vậy \(B=\dfrac{1}{100}\)
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Xem thêm câu trả lời
Cho A = \(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}}\)
A rút gọn bằng bao nhiu????????????
\(\dfrac{2.(1.99+3.97+...+97.3+98.2+99.1)}{1^2+2^2+3^2+...+97^2+98^2+99^2}\)
10 tháng 11 2023 lúc 21:45
dfrac{-1}{20}+dfrac{-1}{30}+dfrac{-1}{42}+dfrac{-1}{56}+dfrac{-1}{72}+dfrac{-1}{90}
dfrac{5}{2.1}+dfrac{4}{1.11}+dfrac{3}{11.2}+dfrac{1}{2.15}+dfrac{13}{15.4}
cíu mình 2 bài này với mình ko biết làm
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\(\dfrac{-1}{20}\)+\(\dfrac{-1}{30}\)+\(\dfrac{-1}{42}\)+\(\dfrac{-1}{56}\)+\(\dfrac{-1}{72}\)+\(\dfrac{-1}{90}\)=
\(\dfrac{5}{2.1}\)+\(\dfrac{4}{1.11}\)+\(\dfrac{3}{11.2}\)+\(\dfrac{1}{2.15}\)+\(\dfrac{13}{15.4}\)=
cíu mình 2 bài này với mình ko biết làm
20 tháng 3 2021 lúc 16:09
Giúp mk vớiCâu 1:Cho A dfrac{1}{dfrac{99}{dfrac{1}{2}+}}+dfrac{2}{dfrac{98}{dfrac{1}{3}+}}+dfrac{3}{dfrac{97}{dfrac{1}{4}+....}}+...+dfrac{99}{dfrac{1}{dfrac{1}{100}}}. B dfrac{92}{dfrac{1}{45}+}-dfrac{1}{dfrac{9}{dfrac{1}{50}+}}-dfrac{2}{dfrac{10}{dfrac{1}{55}+}}-dfrac{3}{dfrac{11}{dfrac{1}{60}+....}}-...dfrac{92}{dfrac{100}{dfrac{1}{500}}}. Tính dfrac{A}{B}
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Giúp mk với
Câu 1:
Cho A = \(\dfrac{1}{\dfrac{99}{\dfrac{1}{2}+}}+\dfrac{2}{\dfrac{98}{\dfrac{1}{3}+}}+\dfrac{3}{\dfrac{97}{\dfrac{1}{4}+....}}+...+\dfrac{99}{\dfrac{1}{\dfrac{1}{100}}}\).
B =\(\dfrac{92}{\dfrac{1}{45}+}-\dfrac{1}{\dfrac{9}{\dfrac{1}{50}+}}-\dfrac{2}{\dfrac{10}{\dfrac{1}{55}+}}-\dfrac{3}{\dfrac{11}{\dfrac{1}{60}+....}}-...\dfrac{92}{\dfrac{100}{\dfrac{1}{500}}}\). Tính \(\dfrac{A}{B}\)