tìm x, biết:
1 - 3 + 3^2 - 3^3 + ... + ( -3 )^x = 9^1013 + 1 / 4
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VH
1
12 tháng 9 2019
Mk sửa 1013 thành 1008 nhá
\(\frac{x-2}{2015}+\frac{x-3}{2014}=\frac{x-1}{1008}\)
\(\Leftrightarrow\frac{x-2}{2015}+\frac{x-3}{2014}-2=\frac{x-1}{1008}-2\)
\(\Leftrightarrow\left(\frac{x-2}{2015}-1\right)+\left(\frac{x-3}{2014}-1\right)=\frac{x-1}{1013}-2\)
\(\Leftrightarrow\frac{x-2-2015}{2015}+\frac{x-3-2014}{2014}=\frac{x-1-2016}{1008}\)
\(\Leftrightarrow\frac{x-2017}{2015}+\frac{x-2017}{2014}=\frac{x-2017}{1008}\)
\(\Leftrightarrow\frac{x-2017}{2015}+\frac{x-2017}{2014}-\frac{x-2017}{1008}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}\right)=0\)
\(\Leftrightarrow x-2017=0\times\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}\right)\)
\(\Leftrightarrow x-2017=0\)
\(\Leftrightarrow x=2017\)
Hok TOT ^_^
D
28 tháng 3 2016
\(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)
\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2016}-1\right)-\left(\frac{x-3}{2014}-1\right)=\left(\frac{x-4}{2013}-1\right)\)
\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}=\frac{x-2017}{2013}\)
\(\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
\(x-2017=0\left(vì\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\right)\)
x=2017
NT
1
TT
16 tháng 11 2023
a) 1/4(x-3)+2=1/5
1/4.(x-3) = 1/5-2
1/4.(x-3) = -9/5
x-3 = (-9/5):1/4
x-3 = -36/5
x = -36/5+3
x= -21/5
2 tháng 7 2023
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
MM
0
LD
18 tháng 7 2017
\(3.\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)
\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)
\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
\(\Rightarrow\)\(x=2012\)
8 tháng 3 2021
a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0
b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
8 tháng 3 2021
c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)
\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)
hay \(x=\dfrac{-1}{9}\)
Vậy: \(x=\dfrac{-1}{9}\)