\(3,\\ a,\Rightarrow x\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\\ b,\Rightarrow x^2\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\\ c,\Rightarrow x^2+5x-6=0\Rightarrow\left(x-1\right)\left(x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\\ d,\Rightarrow\left(x-1\right)\left(x-2014\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2014\end{matrix}\right.\)
\(4,\\ a,2x^3-3x^2+x+a⋮x+2\\ \Rightarrow2x^3-3x^2+x+a=\left(x+2\right)\cdot a\left(x\right)\)
Thay \(x=-2\)
\(\Rightarrow2\left(-8\right)-3\cdot4-2+a=0\\ \Rightarrow-30+a=0\Rightarrow a=30\)
\(b,f\left(x\right)⋮g\left(x\right)\Rightarrow x^2+4x+n=\left(x-2\right)\cdot b\left(x\right)\)
Thay \(x=2\)
\(\Rightarrow4+8+n=0\Rightarrow n=-12\)
\(1,\\ a,x-x^2-1=-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}< 0\)
\(b,f\left(x\right)=x^2-4x+9=\left(x-2\right)^2+5\ge5\\ f\left(x\right)_{max}=5\Leftrightarrow x=2\)
Bài 3:
a: Ta có: \(x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b: Ta có: \(x^3-4x=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
