a/ \(x^3+4x^2+x-6=0\)
\(\Leftrightarrow x^3+3x^2+2x^2+6x-x^2-3x-2x-6=0\)
\(\Leftrightarrow x^2\left(x+3\right)+2x\left(x+3\right)-x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+2x-x-2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[x\left(x+2\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=1\end{matrix}\right.\)
Vậy...........................
b/ \(5-2\left|2x-3\right|=\dfrac{5}{2}x\)(????)
\(\Leftrightarrow-2\left|2x-3\right|=\dfrac{5}{2}x-5\)
\(\Leftrightarrow2\left|2x-3\right|=5-\dfrac{5}{2}x\)
+) Với \(x\ge\dfrac{3}{2}\) ta có:
\(2\left(2x-3\right)=5-\dfrac{5}{2}x\)
\(\Leftrightarrow4x+\dfrac{5}{2}x=5+6\Leftrightarrow\dfrac{13}{2}x=11\Leftrightarrow x=\dfrac{22}{13}\left(tm\right)\)
+) Với \(x\le\dfrac{3}{2}\) có:
\(2\left(2x-3\right)=\dfrac{5}{2}x-5\)
\(\Leftrightarrow4x-\dfrac{5}{2}x=-5+6\)
\(\Leftrightarrow\dfrac{3}{2}x=1\Leftrightarrow x=\dfrac{2}{3}\) (tm)
Vậy pt có 2 nghiệm.................
a) x3+4x2+x-6=0
<=> (x-1)(x+2)(x+3)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
