Tính tổng S=1/1.3+1/2.4+1/3.5+.....+1/4.9+1/8.10
Ai giải giúp mik đi mik phải làm bài kiểm tra để nộp cô.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
S=(1/1.3+1/3.5+.....+1/7.9)+(1/2.4+1/4.8+1/8.10)
2S=1/2.(1-1/3+1/5-1/5+....+1/7-1/9)+(1/2-1/4+1/4-1/8+1/8-1/10)
2S=1/2.(1-1/9)+(1/2-1/10)
2S=1/2.(8/9+2/5)
Ta có: \(2S=\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{7\cdot9}+\dfrac{2}{8\cdot10}\)
\(\Leftrightarrow2S=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{10}\)
\(\Leftrightarrow2S=1+\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{58}{45}\)
\(\Rightarrow S=\dfrac{29}{45}\)
Ta có:
\(S=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}+\dfrac{1}{8.10}\)
\(=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}\right)\) \(+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{8.10}\right)\)
Đặt \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{7.9}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)=\dfrac{4}{9}\)
Đặt \(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{8.10}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{8.10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{5}\)
\(\Rightarrow S=A+B=\dfrac{4}{9}+\dfrac{1}{5}=\dfrac{29}{45}\)
Vậy \(S=\dfrac{29}{45}\)
S=(1/1.3+1/3.5+.....+1/7.9) + (1/2.4+1/4.6+....+1/8.10)
2S=1/2.(1-1/9+(1/2-1/10))
2S=1/2.(8/9 + 2/5)
2S=1/2.58/45
2S=29/45
S=29/45:2
S=29/90
S=(1/1.3+1/3.5+.....+1/7.9) + (1/2.4+1/4.6+....+1/8.10)
2S=1/2.(1-1/9+(1/2-1/10))
2S=1/2.(8/9 + 2/5)
2S=1/2.58/45
2S=29/45
S=29/45:2
S=29/90
S= 1/1.3+ 1/2.4+1/3.5+....+!/7.9+1/8.10
=1/2(1-1/3 +1/2-1/4 +1/3-1/5 +...+ 1/7-1/9 + 1/8-1/10)
=1/2(1+1/2-1/9-1/10)
=....
\(S=\frac{1}{1.3}+.....+\frac{1}{8.10}\)
\(2S=\frac{2}{1.3}+....+\frac{2}{8.10}\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{8}-\frac{1}{10}\)
\(2S=1-\frac{1}{10}\)
\(2S=\frac{9}{10}\)
\(S=\frac{9}{10}:2\)
\(S=\frac{9}{20}\)
S =\(\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+....\frac{1}{2}\left(\frac{1}{8}-\frac{1}{10}\right)\)
S = 1/2 ( 1 -1/3 +1/2-1/4+......+ 1/8-1/10)
S = 1/2(1+1/2-1/9-1/10)
S= 29/45
Bạn nói cô giáo sửa đề thành:
Tính tổng S=1/1.3+1/2.4+1/3.5+.....+1/\(7\).9+1/8.10
chứ không tổng S lẻ lắm, chẳng ai muốn tính cả.