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Đặt A = 1/3+1/15+1/35+1/63+1/99+1/143+1/195
2A= 2/﴾1.3﴿ + 2/﴾3.5﴿ + 2/﴾5.7﴿ + 2/﴾7.9﴿+2/﴾9.11﴿ + 2/﴾11.13﴿+2/﴾13.15﴿
2A=1/1‐1/3+1/3‐1/5+1/5‐1/7+1/7‐1/9+1/9...
2A=1/1‐1/15=14/15
Vậy A=14/15 : 2 = 7/15
A=1/1x3 + 1/3x5 + 1/5x7.......+1/13x15
2A=2/1x3 + 2/3x5 + 2/5x7.......+2/13x15
2A=1-1/3+1/3-1/5+1/5-1/7.......+1/13-1/15
2A=1-1/15
A= 7/15
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1/3 + 1/15 + 1/35+ 1/63 +...... + 1/195
= 1/3 + 1/3x5 + 1/5 x7 + 1/7x9 + ....+1/13x15
= 1/3+1/3-1/5+1/5-1/7+1/7-1/9+....+1/13-1/15 ( vì +- nên rút gọn )
= 1/3+1/3-1/15
=3/5
=1/1.3+1/3.5+1/5.7+...+1/13.15
=1/2.2(1/1.3+1/3.5+1/5.7+...+1/13.15)
=1/2(2/1.3+2/3.5+2/5.7+...+2/13.15)
=1/2(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)
=1/2[(1-1/15)+(1/3-1/3)+(1/5-1/5)+...+(1/13-1/15)]
=1/2[(1-1/15)+0+...+0=1/2(1-1/15)=1/2.14/15=14/30=7/15
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A = \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}+\frac{1}{225}=\)\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}+\frac{1}{225}\)
2A = \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}+\frac{2}{225}=\)\(\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{15-13}{13\cdot15}+\frac{2}{225}\)
2A = \(\frac{5}{3\cdot5}-\frac{3}{3\cdot5}+\frac{7}{5\cdot7}-\frac{5}{5\cdot7}+...+\frac{15}{13\cdot15}-\frac{13}{13\cdot15}+\frac{2}{225}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+\frac{2}{225}\)
2A = \(\frac{1}{3}-\frac{1}{15}+\frac{2}{225}=\frac{60}{225}+\frac{2}{225}=\frac{62}{225}\)
=> A = \(\frac{62}{225}\div2=\frac{31}{225}\)
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Tk mình đi mọi người mình bị âm nè!
ai tk mình mình tk lại cho!!!
( 1/13 + 1/15 + 1/35 + 1/63 + 1/99 ) x X = 2/3
X = 2/3 : ( 1/13 + 1/15 + 1/35 + 1/63 + 1/99 )
X = 286/85
k mình đi mình đang bị âm
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Ta có:1/3=1/1*3;1/15=1/3*5;1/35=1/5*7;1/63=1/7*9.
Ta thấy các phân số trên đều có mẫu số tách được thành các số lẻ liên tiếp và tử số là 1.Số lẻ sau 9 là 11.
Vậy mẫu số của phân số cuối là: 9*11=99
Phân số đó là 1/99
Đáp số : 1/99
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{35}+\frac{1}{48}+\frac{1}{63}+\frac{1}{80}\)
\(=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)
\(=\frac{1}{2}\times\left[\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\right)+\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\frac{2}{8\times10}\right)\right]\)
\(=\frac{1}{2}\times\left[\left(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}\right)+\left(\frac{4-2}{2\times4}+\frac{6-4}{4\times6}+\frac{8-6}{6\times8}+\frac{10-8}{8\times10}\right)\right]\)
\(=\frac{1}{2}\times\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right]\)
\(=\frac{1}{2}\times\left[\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)
\(=\frac{29}{45}\)
\(x:\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\right)=\dfrac{5}{3}\Leftrightarrow x:\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{5}{3}\)
\(\Leftrightarrow x:\dfrac{14}{15}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{5}{3}.\dfrac{14}{15}=\dfrac{14}{9}\)
Đặt A = 1 / 3 + 1 / 15 + 1 / 35 + 1 / 63 + 1 / 99 + 1 / 143 + 1 / 195
A = 1 / 1 x 3 + 1 / 3 x 5 + 1 / 5 x 7 +1 / 7 x 9 + 1 / 9 x 11 + 1 / 11 x 13 + 1 / 13 x 15
A x 2 = 2 / 1 x 3 + 2 / 3 x 5 +2/ 5 x 7 + 2/ 7 x 9 + 2 / 9 x 11 + 2/ 11 x 13 +2 / 13 x 15
A x 2 = 1 / 1 - 1 / 3 + 1 / 3 - 1 /5 + 1 / 5 - 1 / 7 + 1 / 7 - 1 / 9 + 1 / 9 - 1 / 11 + 1 / 11 - 1 / 13 + 1 / 13 - 1 / 15
A x 2 = 1 / 1 - 1 / 15
A x 2 = 14 / 15
A = 7 / 15
=> X : 7/15 = 5/3
=> X = 5/3 x 7/15
=> X = 7/9