$\begin{array}{l}1)\quad \text{Thực hiện phép tính}\\ a)\quad -2x(-3x + 2) - (x+2)^2\\ = 6x^2 - 4x - (x^2 + 4x + 4)\\ = 6x^2 - 4x - x^2 - 4x - 4\\ = 5x^2 - 8x - 4\\ b)\quad (x+2)(x^2 - 2x + 4) - 2(x+1)(1-x)\\ = x^3 + 8 + 2(x^2 - 1)\\ = x^3 + 2x^2 + 6\\ c)\quad (2x - 1)^2 - 2(4x^2 - 1)+ 2(2x+1)^2 \\ = (2x-1)^2 - 2(2x - 1)(2x + 1) + (2x + 1)^2\\ = [(2x - 1) - (2x + 1)]^2\ = (-2)^2 = 4\\ \quad \text{Phân thích đa thức thành nhân tử}\\ a) \quad xy + xz + 3y + 3z\\ = x(y + z) + 3(y+z)\\ = (y+z)(x + 3)\\ b)\quad 4x^2 - 4xy + y^2\\ = (2x)^2 - 2.2x.y + y^2\\ = (2x - y)^2\\ c)\quad x^2 + 2x - 3\\ = x^2 + x - 3x - 3\\ = x(x+1) - 3(x+1)\\ = (x+1)(x-3)\\ 2)\quad A = [(3x-2)(x+1) -(2x-5)(x^2 - 1)]:(x+1)\\ \to A = (3x - 2)(x+1):(x+1) - (2x-5)(x-1)(x+1):(x+1)\\ \to A = 3x - 2 - (2x - 5)(x-1)\\ \to A = 3x - 2 - (2x^2 -7x + 5)\\ \to A = -2x^2 + 4x - 7\\ \text{Với $x=\dfrac12$ ta được:}\\ A = -2\cdot\left(\dfrac12\right)^2 + 4\cdot\dfrac12 - 7\\ \to A = -2\cdot\dfrac14 + 2 - 7\\ \to A = -\dfrac12 - 5\\ \to A = -\dfrac{11}{2}\\ 3a)\quad 6x^2 - (2x - 3)(3x + 2) = 1\\ \Leftrightarrow 6x^2 - (6x^2 + 4x - 9x - 6) -1 =0\\ \Leftrightarrow 5x + 5 = 0\\ \Leftrightarrow x = -1\\ b)\quad (x+1)^3 - (x-1)(x^2 +x+1) -2 =0\\ \Leftrightarrow x^3 + 3x^2 + 3x + 1 - (x^3 - 1) - 2 =0\\ \Leftrightarrow 3x^2 + 3x =0\\ \Leftrightarrow x(x+1) =0\\ \Leftrightarrow \left[\begin{array}{l}x = 0\\x = -1\end{array}\right.\\ c)\quad 2(x-2) = x^2 - 4x + 4\\ \Leftrightarrow 2(x-2)= (x-2)^2\\ \Leftrightarrow (x-2)[(x-2) - 2] = 0\\ \Leftrightarrow (x-2)(x-4) =0\\ \Leftrightarrow \left[\begin{array}{l}x = 2\\x = 4\end{array}\right. \end{array}$