M=3/3.5+3/5.7+3/7.9+...+3/97.99
tính M
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
Tính tổng: M= 3/3.5 + 3/5.7 + 3/7.9 +.......+ 3/95.97 + 3/97.99
Giải:
M=\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{95.97}+\dfrac{3}{97.99}\)
M=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\right)\)
M=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
M=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
M=\(\dfrac{3}{2}.\dfrac{32}{99}\)
M=\(\dfrac{16}{33}\)
Chúc bạn học tốt!
M= 3/3.5 + 3/5.7 + 3/7.9 +.......+ 3/95.97 + 3/97.99
=3.1/2 ( 2/3.5+...+2/97.99)
=3.1/2(1/3- 1/5+...+1/97+1/99)
=3.1/2(1/3- 1/99)
=(3/2).(32/99)
=96/891
n/xét
3/3.5=(3/3-3/5).1/2
3/5.7=(3/5-3/7).1/2
...
3/97.99=(3/97-3/99).1/2
vậy M=(3/3-3/5).1/2+(3/5-3/7).1/2+...+(3/97-3/99).1/2
⇒M=1/2.(3/3-3/7+3/5-3/7+...+3/97-3/99)
=1/2.(3/3-3/99)
=1/2.32/33
M =16/33
VẬY M=16/33
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
c) P = 1/1.2 + 2/2.4 + 3/4.7 + ... + 10/46.56
3/3.5+3/5.7+3/7.9+...+3/49.51 = ?
\(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{49.51}\)
= 3. \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{49}-\frac{1}{51}\right)\)
=\(\frac{3}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
=\(\frac{3}{2}.\frac{16}{51}\)
=\(\frac{8}{17}\)
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Tính nhanh: A= 3/3.5+ 3/5.7+3/7.9 +...+ 3/97.99
\(A=\dfrac{3}{3.5} + \dfrac{3}{5.7} + ... + \dfrac{3}{97.99}\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}.\dfrac{32}{99}\)
\(\Rightarrow A=\dfrac{16}{33}\)
Vậy \(A=\dfrac{16}{33}\)
A= \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{97.99}\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\dfrac{32}{99}\)
= \(\dfrac{3.32}{2.99}\)= \(\dfrac{3.2.3.6}{2.11.3.3}\)= \(\dfrac{6}{11}\)
cho mình hỏi là tại sao lại đặt \(\dfrac{3}{2}\) ra ngoài ?
Tìm x:3/3.5+3/5.7+3/7.9+...+x.(x+2)=8/9
tính hợp lí:
3/3.5 + 3/5.7 + 3/7.9 + ...... + 3/47.49
Đặt A = \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{47.49}\)
2A = \(\dfrac{3.2}{3.5}+\dfrac{3.2}{5.7}+\dfrac{3.2}{7.9}+...+\dfrac{3.2}{47.49}\)
2A = 3\(\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{47.49}\right)\)
2A = 3 \(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
2A = 3 \(\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
2A = 3 . \(\dfrac{46}{147}\)
2A = \(\dfrac{46}{49}\)
=> A = \(\dfrac{46}{49}\) : 2
=> A = \(\dfrac{23}{49}\)
M=2/3.5 + 2/5.7+2/7.9 +....+ 2/97.99
giải thích cho mik đoạn : 1/3-1/5+1/5-1/7
A = 3/4.8/9.15/16....2499/2500
B = 3/3.5 + 3/5.7 + 3/7.9 + .....3/47.49
giúp em với câu này khó quá
B=\(\dfrac{3}{3.5}.\dfrac{3}{5.7}.....\dfrac{3}{47.49}\)
B=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}.\dfrac{2}{5.7}.....\dfrac{2}{47.49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\dfrac{46}{147}\)
B=\(\dfrac{23}{49}\)
a) Ta có: \(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{2499}{2500}\)
\(=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{49\cdot51}{50^2}\)
\(=\dfrac{1}{50}\cdot\dfrac{51}{2}=\dfrac{51}{100}\)
b) Ta có: \(B=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{47\cdot49}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{47\cdot49}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{46}{147}=\dfrac{138}{294}=\dfrac{23}{49}\)