a) \(x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy....
b) \(x\left(x-2\right)\left(x-1\right)\left(x+1\right)=24\)
\(\Leftrightarrow\left(x^2-2x\right)\left(x^2-1\right)-24=0\)
\(\Leftrightarrow x^4-2x^3-x^2+2x-24=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x^2-x+4\right)=0\)
\(\left[{}\begin{matrix}x-3=0\\x+2=0\\x^2-x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Có: \(x^2-x+4=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}>0\)
-> vô nghiệm
Vậy pt có 2 nghiệm....
a) \(x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy..........
