5.7+7.9+9.11+....+25.27
5.7+7.9+9.11+...+25.27
5.7+7.9+9.11+...93.95
Tính:
a) M=2/3.5+2/5.7+2/7.9+...+2/97.99
b) N=3/5.7+3/7.9+3/9.11+...+3/197.199
a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
4/5.7+4/7.9+4/9.11+...+4/53.55 = ?
\(=4\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{55}\right)\)
\(=4.\frac{2}{11}\)
\(=\frac{8}{11}\)
11/5.7+11/7.9+11/9.11+......+11/59.61
3/5.7+3/7.9+3/9.11+...3/59.61
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}\dfrac{2}{7.9}+.........+\dfrac{2}{99.101}\)
\(P=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
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
S=2/5.7+2/7.9+2/9.11+.........+2/93.95
Bg
Ta có: S = \(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
=> S = \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
=> S = \(\frac{1}{5}-\frac{1}{95}\)
=> S = \(\frac{19}{95}-\frac{1}{95}\)
=> S = \(\frac{18}{95}\)
Vậy S = \(\frac{18}{95}\)
\(S=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{93\cdot95}\)
\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(S=\left(\frac{1}{5}-\frac{1}{95}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)+...+\left(\frac{-1}{93}+\frac{1}{93}\right)\)
\(S=\left(\frac{1}{5}-\frac{1}{95}\right)\)
\(S=\frac{19}{95}-\frac{1}{95}\)
\(S=\frac{18}{95}\)
Bài làm :
Ta có:
\(S=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(S=\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}+...+\frac{95-93}{93.95}\)
\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(S=\frac{1}{5}-\frac{1}{95}\)
\(S=\frac{18}{95}\)
\(\text{Vậy S =}\frac{18}{95}\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
N = 3/5.7 + 3/7.9 + 3/9.11 +.....+3/197.199
N=3/2.(1/5.7+1/7.9+1/9.11+....+1/197.199)
N=3/2.(1/5-1/7+1/7-1/9+1/9-1/11+....+1/197-1/199)
N=3/2.(1/5-1/199)
N=3/2.194/995
N=291/995
Có N = 3 . (1/5 . 7 + 1/7 . 9 + .... + 1/197 . 199)
2N = 3 . (2/5 . 7 + 2/7 . 9 + ... + 2/197 . 199)
2N = 3 . (1/5 - 1/7 + 1/7 - 1/9 + ... + 1/197 - 1/199)
2N = 3 . (1/5 - 1/199)
N = \(\frac{3.\left(\frac{1}{5}-\frac{1}{199}\right)}{2}\)
N = (3 . 194/995) : 2
Bạn tự tính nốt nhé chúc bạn học tốt