`D(x)=3x^3+x=0`
`\Leftrightarrow 3x^2*x+x=0`
`\Leftrightarrow x(3x^2+1)=0`
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\3x^2+1=0\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\3x^2=-1\text{(loại)}\end{matrix}\right.\)
Vậy, nghiệm của đa thức là `x=0`
`E(x)=x^2-3x+2=0`
`\Leftrightarrow x^2-2x-x+2=0`
`\Leftrightarrow (x^2-2x)-(x-2)=0`
`\Leftrightarrow x(x-2)-(x-2)=0`
`\Leftrightarrow (x-2)(x-1)=0`
`\Leftrightarrow `\(\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy, nghiệm của đa thức là `x= {2 ; 1}`
`F(x)=4x^2-4x+1=0`
`\Leftrightarrow (2x+1)^2=0`
`\Leftrightarrow 2x-1=0`
`\Leftrightarrow 2x=1`
`\Leftrightarrow x=1/2`
Vậy, nghiệm của đa thức là `x=1/2`
`D(x)=3x^3+x`
`-> 3x^3 +x=0`
`=> x(3x^2 +1)=0`
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x^2+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x^2=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\left\{0\right\}\)
__
`E(x)=x^2-3x+2`
`-> x^2-3x+2=0`
`=> x^2 -2x-x+2=0`
`=> (x^2-2x) -(x-2)=0`
`=> x(x-2)-(x-2)=0`
`=>(x-2)(x-1)=0`
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{2;1\right\}\)
__
`F(x)=4x^2-4x+1`
`-> 4x^2-4x+1=0`
`=> 4x^2 -2x-2x+1=0`
`=> (4x^2-2x)-(2x-1)=0`
`=> 2x(2x-1)-(2x-1)=0`
`=> (2x-1)(2x-1)=0`
`=>(2x-1)^2=0`
`=>2x-1=0`
`=>2x=1`
`=>x=1/2`
Vậy \(x\in\left\{\dfrac{1}{2}\right\}\)
Hoặc
`->4x^2-4x+1=0`
`=> (2x-1)^2=0`
`=> 2x-1=0`
`=>2x=1`
`=>x=1/2`
Vậy \(x\in\left\{\dfrac{1}{2}\right\}\)
