Question

Giúp em vs

Answer 1

Bài 3:

\(a,\Leftrightarrow x^2-4x+4-x^2-4x=0\\ \Leftrightarrow8x=4\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\\ \Leftrightarrow4\left(x+2\right)=0\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Bài 4:

\(a,A:B=\left(x^3-3x^2-2x^2+6x+6x-18+9\right):\left(x-3\right)\\ =\left[\left(x-3\right)\left(x^2-2x+6\right)+9\right]:\left(x-3\right)\\ =x^2-2x+6\left(dư.9\right)\\ b,A⋮B\Leftrightarrow9⋮x-3\\ \Leftrightarrow x-3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;4;6;12\right\}\)