\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+...+\dfrac{9997}{9999}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NP
1
H9
HT.Phong (9A5)
CTVHS
1 tháng 4 2023
\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+\dfrac{61}{63}+\dfrac{97}{99}\)
\(=\left(1-\dfrac{2}{3}\right)+\left(1-\dfrac{2}{15}\right)+\left(1-\dfrac{2}{35}\right)+\left(1-\dfrac{2}{63}\right)+\left(1-\dfrac{2}{99}\right)\)
\(=\left(1+1+1+1+\right)-\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}\right)\)
\(=5-\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)\)
\(=5-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=5-\left(1-\dfrac{1}{11}\right)\)
\(=5-\dfrac{10}{11}\)
\(=\dfrac{45}{11}\)
11 tháng 3 2021
1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)
\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)
\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))
\(\Leftrightarrow x=-36\).
Vậy nghiệm của pt là x = -36.
17 tháng 7
2) x(x+1)(x+2)(x+3)= 24
⇔ x.(x+3) . (x+2).(x+1) = 24
⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24
Đặt \(x^2\)+ 3x = b
⇒ b . (b+2)= 24
Hay: \(b^2\) +2b = 24
⇔\(b^2\) + 2b + 1 = 25
⇔\(\left(b+1\right)^2\)= 25
+ Xét b+1 = 5 ⇒ b=4 ⇒ \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0
⇒(x-1)(x+4)=0⇒x=1 và x=-4
+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0
⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))2 = - \(\dfrac{15}{4}\) Hay ( \(x^2\) +\(\dfrac{3}{2}\) )2= -\(\dfrac{15}{4}\) (vô lí)
⇒x= 1 và x= 4
MH
2
28 tháng 1
\(A=\dfrac{6}{3}+\dfrac{6}{15}+\dfrac{6}{35}+...+\dfrac{6}{9999}\)
\(=3\left(\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\right)\)
\(=3\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=3\left(1-\dfrac{1}{101}\right)=3\cdot\dfrac{100}{101}=\dfrac{300}{101}\)
PJ
2
Q
5 tháng 4 2017
\(A=\dfrac{-1}{3}+\dfrac{-1}{15}+\dfrac{-1}{35}+...+\dfrac{-1}{9999}\)
\(\Rightarrow-A=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\)
\(-A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(-2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(-2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(-2A=1-\dfrac{1}{101}\)
\(-2A=\dfrac{100}{101}\)
\(-A=\dfrac{100}{101}:2\)
\(-A=\dfrac{50}{101}\)
\(\Rightarrow A=\dfrac{-50}{101}\)
Chúc bạn học tốt!
BN
5 tháng 4 2017
\(A=\dfrac{-1}{3}+\dfrac{-1}{15}+\dfrac{-1}{35}+...+\dfrac{-1}{9999}\)
\(A=-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\)
Đặt \(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+....+\dfrac{1}{9999}\)
\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(2B=1-\dfrac{1}{101}=\dfrac{100}{101}\)
\(B=\dfrac{100}{101}:2=\dfrac{50}{101}\)
\(\Rightarrow A=-B=-\dfrac{50}{101}\)
HT
0
NN
1
12 tháng 1 2019
1/3+13/15+33/35+31/63+.....................+9601/9603+9997/9999
\(=1-\frac{2}{3}+1-\frac{2}{15}+...+1-\frac{2}{9999}\)
\(=\left(1+1+1+1+...+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)
\(=50-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{4950}{101}\)
HTDT
\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+...+\dfrac{9997}{9999}\)
\(=1-\dfrac{2}{3}+1-\dfrac{2}{15}+1-\dfrac{2}{35}+...+1-\dfrac{2}{9999}\)
\(=\left(1+1+1+...+1\right)-\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\)
\(=50-1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=50-\left(1-\dfrac{1}{101}\right)=50-\dfrac{100}{101}\)
\(=\dfrac{4950}{101}\)