Tìm x:
x-2=(x-2)2
(x2+3).(x+1)+x=-1
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\(\Leftrightarrow x^3+2x^2-3x-x^3-3x^2=-4\)
\(\Leftrightarrow x^2+3x-4=0\)
=>(x+4)(x-1)=0
=>x=-4 hoặc x=1
1) \(x:\dfrac{1}{3}=\dfrac{1}{2}+\dfrac{1}{3}\)
\(\Rightarrow3\times x=\dfrac{5}{6}\Rightarrow x=\dfrac{5}{18}\)
2) \(\left(x:\dfrac{2}{3}\right):\dfrac{2}{5}=\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{2}\times x=\dfrac{7}{10}\times\dfrac{2}{5}=\dfrac{7}{25}\)
\(\Rightarrow x=\dfrac{7}{25}:\dfrac{3}{2}=\dfrac{14}{75}\)
\(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)^2=8.2\)
=>(x+1)2=16
=>x+1=\(\sqrt{16}\)
=>x+1=4
=>x=3
\(6x\left(1-3x\right)+9x\left(2x-7\right)+171=0\)
\(\Leftrightarrow6x-18x^2+18x^2-63x+171=0\)
\(\Leftrightarrow-57x=-171\)
\(\Leftrightarrow x=3\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)-\left(\frac{x+3}{2013}+1\right)-\left(\frac{x+4}{2012}+1\right)=0\)
\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2016=0\) ( vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\) )
\(\Leftrightarrow x=-2016\)
\(x+\left(x+1\right)+\left(x+2\right)+...+19+20=20\)
\(\Leftrightarrow x+\left(x+1\right)+\left(x+2\right)+...+19=0\)
\(\Leftrightarrow\frac{x\left(19+x\right)}{2}=0\)
\(\Leftrightarrow x\left(x+19\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+19=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-19\end{cases}}\)
Mà \(x\ne0\)nên \(x=-19\)
\(x-2=\left(x-2\right)^2\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\in\left\{2,3\right\}\)
\(\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+4\right)=0\)
\(\Rightarrow x=-1\)