Cau 2:
\(a,=\left(x^2-4\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\left(x+1\right)\\ b,=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ c,=\left(x^2+6x+9\right)\left(x^2-6x+9\right)=\left(x-3\right)^2\left(x+3\right)^2\\ d,=25-\left(x-y\right)^2=\left(25-x+y\right)\left(25+x-y\right)\\ e,=x^3-x^2-3x^2+3x+x-1=\left(x-1\right)\left(x^2-3x+1\right)\\ f,=3\left(x-y\right)-\left(x-y\right)^2=\left(3-x+y\right)\left(x-y\right)\)
Cau 3:
\(a,\Rightarrow\left(x-3\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\\ b,\Rightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ c,\Rightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\\ d,\Rightarrow x^3-2x^2-2x+4=0\\ \Rightarrow\left(x-2\right)\left(x^2-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\\ e,\Rightarrow8x^3+4x^2-12x^2-6x+2x+1=0\\ \Rightarrow\left(2x+1\right)\left(4x^2-6x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3\pm\sqrt{5}}{4}\end{matrix}\right.\)
