tìm x:
x(x-3)-3x+9=0
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\(6x\left(1-3x\right)+9x\left(2x-7\right)+171=0\)
\(\Leftrightarrow6x-18x^2+18x^2-63x+171=0\)
\(\Leftrightarrow-57x=-171\)
\(\Leftrightarrow x=3\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)-\left(\frac{x+3}{2013}+1\right)-\left(\frac{x+4}{2012}+1\right)=0\)
\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2016=0\) ( vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\) )
\(\Leftrightarrow x=-2016\)
=>x(x^2-9/16)=0
=>x(x-3/4)(x+3/4)=0
=>x=0; x=3/4; x=-3/4
a: \(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)
hay \(x\in\left\{0;\sqrt{3};-\sqrt{3}\right\}\)
b: \(=\dfrac{x^3-3x^2+6x-8}{x-2}=\dfrac{x^2-2x-x^2+2x+4x-8}{x-2}=x^2-x+4\)
\(x=\dfrac{4}{5}\times\dfrac{4}{3}\)
\(x=\dfrac{16}{15}\)
-----------------------
\(x=\dfrac{5}{9}\times\dfrac{3}{8}\)
\(x=\dfrac{5}{24}\)
\(X:\dfrac{3}{9}=\dfrac{7}{8}\)
\(X=\dfrac{7}{8}\times\dfrac{3}{9}\)
\(X=\dfrac{21}{72}\)
\(\left(X-21\times13\right):11=30\)
\(X-273=330\)
\(X=603\)
`x:3/9=7/8`
`x=7/8xx3/9`
`x=7/24`
`-------------`
`(x-21xx13):11=30`
`(x-273):11=30`
`x-273=30xx11`
`x-273=330`
`x=330-273`
`x=57`
Theo đề ta có:\(x.x=x:x=0+x=x\)
\(\Rightarrow x^2=x:x=0+x\)
Xét theo vế \(x^2=1\)thì ta sẽ có 3 x là -1 ;0 ;1
Xét theo vế \(x:x=x=>1=x\)=> x là 1
Vì 1 thỏa mãn điều kiện bài cho nên x=1
\(x\left(x-3\right)-3x+9=0\)
\(x\left(x-3\right)-3\left(x-3\right)=0\)
\(\left(x-3\right)\left(x-3\right)=0\)
\(\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy x = 3
(x • (x - 3) - 3x) + 9 = 0
(x - 3)2 = 0
(x-3)2 = 0
: x-3 = 0
x = 3
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