\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
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TL
1
IA
9 tháng 6 2018
\(\Leftrightarrow10\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{x\times\left(x+1\right)}\right)=9\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=9\div10\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{9}{10}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{10}\)
\(\Rightarrow x+1=10\)
\(\Leftrightarrow x=9\)
Vậy x = 9
2 tháng 4 2019
Ở link này có bài tham khảo nè bn :
http://olm.vn/hoi-dap/detail/42438427638.html
2 tháng 6 2015
( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +1/5.6 ) x 10 - x = 0
= ( 1- 1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 - 1/5 +1/5 -1/6 ) x 10 - x = 0
= ( 1 - 1/6 ) x 10 - x = 0
= 5/6 x 10 - x =0
= 25/3 - x =0
x = 25/3 - 0
x = 25/3
MT
2 tháng 6 2015
\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)\times10-x=0\)
\(\frac{5}{6}\times10-x=0\)
\(\frac{25}{3}-x=0\)
x =\(\frac{25}{3}-0=\frac{25}{3}\)
22 tháng 1 2017
Ta thấy :
1/1x2 = 1/1 - 1/2
1/2x3 = 1/2 - 1/3
....
=>( 1/1x2 + 1/2x3 + 1/3x4 + 1/5x6 ) x 10 - x = ( 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 ) x 10 - x
= ( 1/1 - 1/6 ) x 10 - x =0
5/6 x 10 - x = 0
25/3 - x = 0
=> x = 25/3
YY
24 tháng 4 2018
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{9}-\frac{1}{10}\)
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}\)
\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)
\(x=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}\)
PY
24 tháng 4 2018
Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)
= \(1-\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
16 tháng 8 2015
Ta co:
\(\left[\left(\frac{29}{25}-x\right).\frac{21}{4}\right]:\left[\left(\frac{95}{9}-\frac{29}{4}\right).\frac{36}{17}\right]=\frac{3}{4}\)
\(\left[\left(\frac{29}{25}-x\right).\frac{21}{4}\right]:\left(\frac{119}{36}.\frac{36}{17}\right)=\frac{3}{4}\)
\(\left[\left(\frac{29}{25}-x\right).\frac{21}{4}\right]:7=\frac{3}{4}\)
\(\left(\frac{29}{25}-x\right).\frac{21}{4}=\frac{3}{4}.7=\frac{21}{4}\)
\(\frac{29}{25}-x=\frac{21}{4}:\frac{21}{4}=1\)
\(x=\frac{29}{25}-1=\frac{4}{25}\)
26 tháng 2 2020
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\left(x\ne1\right)\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x}{x^2+x+1}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\left(x^2+x+1-3x^2-2x^2+2x\right)=0\)
\(\Leftrightarrow-4x^2+3x+1=0\left(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\ne0\right)\)
\(\Leftrightarrow-4x^2+4x-x+1=0\)
\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\-4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\-4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\left(loại\right)\\x=\frac{-1}{4}\end{cases}}}\)
Vậy \(x=\frac{-1}{4}\)
6 tháng 8 2016
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{x\left(x+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=1-\frac{1}{x+1}\)
\(=\frac{x+1-1}{x+1}=\frac{x}{x+1}\)
6 tháng 8 2016
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x.\left(x+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=1-\frac{1}{x+1}\)
\(=\frac{x+1}{x+1}-\frac{1}{x+1}\)
\(=\frac{x}{x+1}\)
<=> (1-1/10)(x-1)+x/10=x-9/10
<=> 9x/10-9/10+x/10=x-9/10
<=> x=x
Như vậy, phương trình thỏa mãn với mọi x