\(1,\left(A+B\right)^2=A^2+2AB+B^2\\ \left(A-B\right)^2=A^2-2AB+B^2\\ A^2-B^2=\left(A-B\right)\left(A+B\right)\\ 2,\\ A,=2x^2y^2\left(2x+3-4x^2y\right)\\ b,=\left(2x-3y\right)\left(2x+3y\right)+\left(2x-3y\right)=\left(2x-3y\right)\left(2x+3y+1\right)\\ c,=x^3\left(x+5\right)-8\left(x+5\right)=\left(x^3-8\right)\left(x+5\right)\\ =\left(x-2\right)\left(x^2+2x+4\right)\left(x+5\right)\\ d,=3\left(x^2-2x-4y^2+1\right)=3\left[\left(x-1\right)^2-4y^2\right]\\ =3\left(x-1-2y\right)\left(x-1+2y\right)\\ 3,\\ a,\Rightarrow3x^2-3x+6-2x-3x^2=0\\ \Rightarrow-5x=-6\Rightarrow x=\dfrac{6}{5}\\ b,\Rightarrow\left(x-1\right)\left(x-1+x+2\right)=0\\ \Rightarrow\left(x-1\right)\left(2x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(c,\Rightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\\ \Rightarrow\left(x^2+1\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\2x+3=0\end{matrix}\right.\Rightarrow x=-\dfrac{3}{2}\\ d,\Rightarrow2x^2+4x-3x-6=0\\ \Rightarrow2x\left(x+2\right)-3\left(x+2\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\\ 4,A=\left(2x+1\right)\left(x-3\right)=2x^2-5x-3\\ A=2\left(x^2-2\cdot\dfrac{5}{4}x+\dfrac{25}{16}\right)-\dfrac{49}{8}\\ A=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{49}{8}\ge-\dfrac{49}{8}\\ A_{min}=-\dfrac{49}{8}\Leftrightarrow x=\dfrac{5}{4}\)
