Số tự nhiên x thỏa mãn:
B = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)
là x = .....
Trình bày cách giải cụ thể ra nha!
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16 tháng 9 2020
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{7}\)
\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{16}{7}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{7}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{7}\)
\(\Rightarrow\frac{1}{x+2}=-\frac{9}{7}\)
\(\Rightarrow-9\left(x+2\right)=7\)
\(\Rightarrow x+2=-\frac{7}{9}\)
\(\Rightarrow x=-\frac{25}{9}\)
Vậy \(x=-\frac{25}{9}\)
D
18 tháng 2 2016
ta nhân vế trái vs 2:
\(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{x\left(x+2\right)}=\frac{8}{17}\)
\(\frac{1}{ }-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{8}{17}\)
\(1-\frac{1}{x+2}=\frac{8}{17}\)
\(\Rightarrow17\left(x+1\right)=8\left(x+2\right)\)
\(\Rightarrow17x+17=8x+16\)
\(\Rightarrow17x-8x=-17+16\)
\(\Rightarrow9x=-1\)
\(\Rightarrow x=\frac{-1}{9}\)
HA
18 tháng 2 2016
2(1/1.3+1/3.5+1/5.7+...+1/x(x+2) )=16/34 *2
2/1.3+2/3.5+2/5.7+...+2/x(x+2)=32/34=16/17
1/1-1/3+1/3-1/5+1/5-1/7+...+1/x-1/x+2=16/17
1/1-1/x+2=16/17
1/x+2=1/1-16/17
1/x+2=1/17
suy ra x+2=17
x=17=2=15
16 tháng 12 2016
1/2[2/1.3+2/3.5+2/5.7+.........+2/x(x+2)]=16/34
2/1.3+2/3.5+2/5.7+......+2/x(x+2)=16/34:1/2=16/17
1/1-1/3+1/3-1/5+1/5-1/7+.....+1/x-1/x+2=16/17
1-1/x+2=16/17
1/x+2=1-16/17=1/17
suy ra:x+2=17
x=17-2
x=15
29 tháng 12 2016
A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)
\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)
\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)
X=16
10 tháng 1 2016
bài 1
[(x+2)/1010]+ [(x+2)/1111]= [(x+2)/1212]+[(x+2)/1313]
=>[(x+2)/1010]+[(x+2)/1111] - [(x+2)/1212]-[(x+2)/1313] = 0
=>(x+2).[(1/1010)+(1/1111)-(1/1212)-(1/1313)=0
Vì [(1/1010)+(1/1111)-(1/1212)-(1/1313)] khác 0
=>x+2=0
=>x=-2
13 tháng 1 2019
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\left(1-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\frac{14}{15}\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow\frac{14}{15}x-\frac{14}{15}=\frac{6}{5}x-\frac{14}{15}\)
\(\Leftrightarrow-\frac{4}{15}x=\frac{28}{15}\)
\(\Leftrightarrow x=7\)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)
\(\Rightarrow\frac{1}{2}\left[\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}\right]=\frac{16}{17}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{17}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{17}\Rightarrow x+2=17\Rightarrow x=15\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)
Ta có:
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{8}{17}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{17}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{17}\)
\(\Rightarrow x+2=17\)
\(\Rightarrow x=15\)
Vậy \(x=15\)