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????????????
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H9
HT.Phong (9A5)
CTVHS
20 tháng 8 2023
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{97}+\dfrac{1}{3}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{\left(\dfrac{1}{1\cdot99}+\dfrac{1}{99\cdot1}\right)+\left(\dfrac{1}{97\cdot3}+\dfrac{1}{97\cdot3}\right)+...+\left(\dfrac{1}{51\cdot49}+\dfrac{1}{49\cdot51}\right)}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{97\cdot3}+...+\dfrac{100}{49\cdot51}}{\dfrac{2}{1\cdot99}+\dfrac{2}{97\cdot3}+...+\dfrac{2}{51\cdot49}}\)
\(=\dfrac{100\cdot\left(\dfrac{1}{99}+\dfrac{1}{97\cdot3}+...+\dfrac{1}{49\cdot51}\right)}{2\cdot\left(\dfrac{1}{99}+\dfrac{1}{97\cdot3}+...+\dfrac{1}{49\cdot51}\right)}\)
\(=\dfrac{100}{2}\)
\(=50\)
Q
1 tháng 4 2017
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:
\(Q=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\\ =\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{50}}\\ =50\)
HD
1 tháng 4 2017
@Nguyễn Huy Tú, @Hoàng Thị Ngọc Anh, @Tuấn Anh Phan Nguyễn, @Hoang Hung Quan, @ngonhuminh, và các bn khác giúp mk với!!
LB
1 tháng 4 2018
\(Q=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2}{1.99}+\dfrac{2}{3.97}+...+\dfrac{2}{51.49}}\)
\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...+\dfrac{100}{51.49}}\)
\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{\dfrac{1+99}{1.99}+\dfrac{3+97}{3.97}+...+\dfrac{51+49}{51.49}}\)
\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{\dfrac{1}{99}+1+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{51}+\dfrac{1}{49}}\)
\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{1+\dfrac{1}{3}+...+\dfrac{1}{99}}\)
\(\Rightarrow Q=50\)
Q
12 tháng 4 2017
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: \(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{496.501}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{496.501}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}.\dfrac{500}{501}\)
\(=\dfrac{100}{501}\)
Q
12 tháng 4 2017
Bài 2: Tính:
a) \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(A=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(A=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(A=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(\Rightarrow A=\dfrac{100}{2}=50\)
HA
12 tháng 8 2017
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) => \(A:B=\dfrac{100}{2}=50\)
LD
21 tháng 1 2019
Ta có:
\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\dfrac{99+1}{1\cdot99}+\dfrac{97+3}{3\cdot97}+...+\dfrac{1+99}{99\cdot1}}{100}}\)
\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{99}+1\right)}{100}}\)
\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{100}}=\dfrac{1}{\dfrac{2}{100}}=\dfrac{100}{2}=50\)
LD
21 tháng 1 2019
\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+...+\dfrac{1}{99}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{1+\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{100}{100}+\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4 2023
b, B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2 \(\times\) B + B = 1 - \(\dfrac{1}{2^{100}}\)
3B = ( 1 - \(\dfrac{1}{2^{100}}\))
B = ( 1 - \(\dfrac{1}{2^{100}}\)) : 3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4 2023
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
A\(\times\) 3 = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+ \(\dfrac{1}{3^{n-1}}\)
A \(\times\) 3 - A = 3 - \(\dfrac{1}{3^n}\)
2A = 3 - \(\dfrac{1}{3^n}\)
A = ( 3 - \(\dfrac{1}{3^n}\)) : 2
LT
3
29 tháng 4 2015
Ta thấy:
1/1 + 1/99 = (99+1)/(1.99)=100/(1.99)
1/3 + 1/97 = (97+3)/(3.97)=100/(3.97)
1/5 + 1/95 = (95+5)/(5.95)=100/(3.97)
…
1/97 + 1/3 = (3+97)/(97.3)=100/(97.3)
1/99 + 1/1 = (1+99)/(99.1)=100/(99.1)
=>
1/(1.99)=(1/1+1/99)/100
1/(3.97)=(1/3+1/97)/100
…
1/(99.1)=(1/99+1/1)/100
------------------------------ cộng 2 vế của các đẳng thức trên. Ta được đẳng thức:
1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 )
=[(1/1+1/99)+(1/3+1/99)+…+(1/99+1/1)]/1...
=2(1+1/3+1/5+1/7…+1/99]/100
=(1+1/3+1/5+1/7…+1/99]/50
Vậy:
A=(1+1/3+1/5+1/7+...+1/97+1/99) / [ 1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 ) ]
A=(1+1/3+1/5+1/7+...+1/97+1/99)/[(1+1/3...
A=50.
ko chắc nhé
AT
11 tháng 6 2021
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)
11 tháng 6 2021
cả 2 ý bạn trục căn thức ở mấu là xong nhé:
vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy
=100 nha bn
sử dụng shipma trên casio nha bn