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\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{13\cdot15}\)
\(=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\frac{4}{15}\)
\(=\frac{2}{15}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{1}{13.15}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\)4/15
=2/15
Đặt :
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+.........+\dfrac{1}{99.101}\)
\(\Leftrightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+............+\dfrac{2}{99.101}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+............+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{101}\)
\(\Leftrightarrow2A=\dfrac{98}{303}\)
\(\Leftrightarrow A=\dfrac{49}{303}\)
=3(1/1.3+1/3.5+1/5.7+1/7.9)
=3/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) vi khoang cach tu 1-3;3-5;5-7;7-9 la 2 nen ta nhan tat ca voi 1/2 ma 3.1/2=3/2
=3/2.(1-1/9) rut gon -1/3+1/3;-1/5+1/5;-1/7+1/7=0
=3/2.8/9=4/3
ta có :3/(1.3)+3/(3.5)+3/(5.7)+3/(7.9)
ta đặt 3 làm chung rồi tự làm đc
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{87.89}\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{87}-\frac{1}{89}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{89}\right)\)
= \(\frac{1}{2}.\frac{86}{267}=\frac{43}{267}\)
~~~
Đáp số to quá, tớ không chắc là mình đúng đâu.
#Sunrise
Gọi biểu thức trên là A.
Ta có : 2A = \(\frac{2}{3.5}+\frac{2}{5.7}+................+\frac{2}{89.91}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.............+\frac{1}{89}-\frac{1}{91}\)
=> 2A = \(\frac{1}{3}-\frac{1}{91}=\frac{88}{273}\)
=> A = \(\frac{88}{273}\div2=\frac{44}{273}\)
k cho mình nha bạn
Goi tong tren la : A
Ta co: A = 1/3.5 + 1/5.7 + 1/7.9 + ......+ 1/89.91
2A = 2/3.5 + 2/5.7 +......+ 2/89.91
2A = 1/3 - 1/5 + 1/5 - 1/7 +......+ 1/89 - 1/91
2A = 1/3 - 1/91
2A = 88/273
A = 44/273
\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(M=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99})\)
\(M=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(M=2.\dfrac{32}{99}\)
\(M=\dfrac{64}{99}\)
http://vietjack.com/giai-sach-bai-tap-toan-6/bai-95-trang-28-sach-bai-tap-toan-6-tap-2.jsp
A= 1/3-1/5+1/5-1/7+....+1/25-1/27
=1/3-1/27
=8/27
vậy A=8/27
\(Q=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\)\(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)
\(Q=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(Q=\frac{1}{3}+0+0+0-\frac{1}{11}\)
\(Q=\frac{11}{33}-\frac{3}{33}=\frac{8}{33}\)
36
trả lời chi tiết
= 398492859.823050309
dễ lắm bn ơi nhớ tick cho mk nha:
Đặt A = 3.5 + 5.7 + 7.9 + ... + 118.119-
- Nhân 6 vào 2 vế:
-
- Biến đổi:
6A = 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + 7.9.11 - 5.7.9 + ... + 118.119.121 - 117.118.119 6A = -1.3.5 + 118.119.121 6A = -15 + 118.119.121 6A = 118.119.121 - 156A = 3.5.6 + 5.7.6 + 7.9.6 + ... + 118.119.6
6A = 3.5.(7-1) + 5.7.(9-3) + 7.9.(11-5) + ... + 118.119.(121-117)
-
- A = (118.119.121 - 15)/6
Kết quả:A = 6118.119.121−15
=283178
Sửa đề: \(3\cdot5+5\cdot7+7\cdot9+\cdots+117\cdot119\)
Đặt \(A=3\cdot5+5\cdot7+7\cdot9+\cdots+117\cdot119\)
\(=3\left(3+2\right)+5\left(5+2\right)+\cdots+117\left(117+2\right)\)
\(=\left(3^2+5^2+\cdots+117^2\right)+2\left(3+5+\cdots+117\right)\)
\(=\left(1^2+3^2+\cdots+117^2\right)+2\left(1+3+5+\cdots+117\right)-1-2\)
\(=\left(1^2+3^2+\cdots+117^2\right)+2\left(1+3+5+\cdots+117\right)-3\)
Đặt \(B=1^2+3^2+\cdots+117^2\)
\(=1^2+2^2+3^2+4^2+\ldots+118^2-\left(2^2+4^2+\cdots+118^2\right)\)
\(=\left(1^2+2^2+\cdots+118^2\right)-4\left(1^2+2^2+\cdots+59^2\right)\)
\(=\frac{118\left(118+1\right)\left(2\cdot118+1\right)}{6}-4\cdot\frac{59\left(59+1\right)\left(2\cdot59+1\right)}{6}\)
\(=\frac{118\cdot119\cdot237}{6}-4\cdot\frac{59\cdot60\cdot119}{6}=59\cdot119\cdot79-4\cdot59\cdot10\cdot119\)
\(\) \(=59\cdot119\left(79-4\cdot10\right)=59\cdot119\cdot39\)
=273819
Đặt C=1+3+5+...+117
Số số hạng của dãy số là: \(\frac{117-1}{2}+1=\frac{116}{2}+1=58+1=59\) (số)
Tổng của dãy số là: \(\left(117+1\right)\cdot\frac{59}{2}=59\cdot59=3481\)
Ta có: \(A=\left(1^2+3^2+\cdots+117^2\right)+2\left(1+3+5+\cdots+117\right)-3\)
=273819+2*3481-3
=280778