6/5x7+ 6/7x9+...+ 6/95x97+ 6/97x99
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: A = \(\frac{6}{5\times7}+\frac{6}{7\times9}+\frac{6}{9\times11}+...+\frac{6}{95\times97}+\frac{6}{97\times99}\)
\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{95\times97}+\frac{1}{97\times99}\right)\)
\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5}-\frac{1}{99}\right)\)
=> A = ...
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=1x3+3x5+5x7+7x9+...+95x97+97x99
= 1.(1+2)+3.(3+2)+5.(5+2)+....+95.(95+2)+97.(97+2)
= 12+1.2+32+3.2 +52+5.2+...+952+95.2+ 972+97.2
= (12+32 +52+...+952+ 972)+(1.2+3.2 +5.2+...+95.2+97.2)
= (12+32 +52+...+952+ 972)+ 2.(1+3 +5+...+95+97)
Đặt : A = 12+32 +52+...+952+ 972
C =1+3 +5+...+95+97
tính A và C (tìm câu hỏi tương tự hình như anh thấy họ làm rồi đấy) sau đó thay vào tính B
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(6B=1\times3\times6+3\times5\times6+...+97\times99\times6\)
\(=1\times3\times\left(5+1\right)+3\times5\times\left(7-1\right)+5\times7\times\left(9-3\right)+...+97\times99\times\left(101-95\right)\)
\(=1\times3\times5+1.3+3\times5\times7-3\times5\times1+...-97\times99\times95\)
\(=97\times99\times101+3\)
\(\Rightarrow B=\frac{97\times99\times101+3}{6}=161651\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b)
S2=6/2x5+6/5x8+6/8x11+...+6/29x32
=2.(3/2.5+3/5.8+...+3/29.32)
=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)
=2.(1/2-1/32)
=2.15/32
=15/16
a)
Ta có:
S1=2/3x5+2/5x7+2/7x9+...+2/97x99
=1/3-1/5+1/5-1/7+...+1/97-1/99
=1/3-1/99
=32/99
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{6}{5x7}+\frac{6}{7x9}+...\frac{6}{97x99}\)
\(=3x\left(\frac{2}{5x7}+\frac{2}{7x9}+...\frac{2}{97x99}\right)\)
\(=3x\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3x\left(\frac{1}{5}-\frac{1}{99}\right)\)
\(=3x\left(\frac{99}{495}-\frac{5}{495}\right)\)
\(=3x\frac{94}{495}=\frac{94}{165}\)
Vậy \(A=\frac{94}{165}\)
\(\frac{6}{5}\)x 7 + \(\frac{6}{7}\)x 9 + .... + \(\frac{6}{97}\)x 99
= \(\frac{6}{5}\) - \(\frac{6}{7}\)+\(\frac{6}{7}\)- \(\frac{6}{9}\)+ ..... + \(\frac{6}{97}\)- \(\frac{6}{99}\)
= \(\frac{6}{5}\) - \(\frac{6}{99}\)
= \(\frac{188}{165}\)
nhớ cho đúng đó
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{6}{3\cdot5}+\dfrac{6}{5\cdot7}+\dfrac{6}{7\cdot9}+.....+\dfrac{6}{33\cdot35}\)
\(=\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{33\cdot35}\right)\cdot3\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{33}-\dfrac{1}{35}\right)\cdot3\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{35}\right)\cdot3\)
\(=\dfrac{32}{3\cdot35}\cdot3\)
\(=\dfrac{32}{35}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{6}{5.7}+\dfrac{6}{7.9}+...+\dfrac{6}{59.61}\)
\(=3\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
\(=3\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=3\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(=\dfrac{3.56}{305}\\ =\dfrac{168}{305}\)
\(\frac{6}{5\cdot7}+\frac{6}{7\cdot9}+...+\frac{6}{97\cdot99}\)
\(=3\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{97\cdot99}\right)\)
\(=3\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{5}-\frac{1}{99}\right)\)
\(=\)
Cái này là tự uận hay trắc nghiệm . Để mik còn bt và m