a) \(x^3-4x=0\)
\(\Leftrightarrow x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
c) \(x^3+x^2+2x=0\)
\(\Leftrightarrow x\left(x^2+x+2\right)=0\)
\(\Leftrightarrow x\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+x+2=0\left(vo-nghiem\right)\end{matrix}\right.\)
d) \(2x^2-5x+3=0\)
\(\Leftrightarrow2x^2-2x-3x+3=0\)
\(\Leftrightarrow2x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{2}\end{matrix}\right.\)
phân tích đa thức thành nhân tử chứ?
a) x3-4x <=> x(x2-4) <=> x(x-2)(x+2)
b) x2-4 <=> (x-2)(x+2)
c) x3+x2+2x <=> x(x2+x+2)
d) 2x2-5x+3
<=> 2x2-2x-3x+3
<=> 2x(x-1)-3(x+1)
<=> làm sao chuyển dấu :v qên r
\(x^3-4x=0\Leftrightarrow x\left(x^2-4\right)=0\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=-2\end{matrix}\right.\)
\(x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
\(x^3+x^2+2x=0\Leftrightarrow x\left(x^2+x+2\right)=0\text{ dê thấy: }x^2+x+2>0\Rightarrow x=0\)
\(2x^2-5x+3=0\Leftrightarrow x^2-\frac{5}{2}x+\frac{3}{2}=0\Leftrightarrow x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{1}{16}=0\Leftrightarrow\left(x-\frac{5}{4}\right)^2=\frac{1}{16}\Leftrightarrow x-\frac{5}{4}=\pm\frac{1}{4}\Leftrightarrow x\in\left\{1;\frac{3}{2}\right\}\)
