解：移项，得$3(x - 3)-(x - 3)^2 = 0.$ 提取公因式，得$(x - 3)(3 - x + 3)=0.$ 所以$x - 3 = 0$或$3 - x + 3 = 0.$ 解得$x_1 = 3,x_2 = 6$
解：对于方程$x^2 - 5x + 6 = 0，$因式分解得$(x - 2)(x - 3)=0，$则$x - 2 = 0$或$x - 3 = 0，$解得$x_1 = 2,x_2 = 3$
解：对于方程$x(x + 4)-x - 4 = 0，$变形为$x(x + 4)-(x + 4)=0，$提取公因式得$(x + 4)(x - 1)=0，$则$x + 4 = 0$或$x - 1 = 0，$解得$x_1 = 1,x_2 = -4$
解：对于方程$10x^2 - 3x-\frac{1}{5}=x^2 - 3x+\frac{4}{5}，$移项合并同类项得$9x^2 - 1 = 0，$因式分解得$(3x + 1)(3x - 1)=0，$则$3x + 1 = 0$或$3x - 1 = 0，$解得$x_1=\frac{1}{3},x_2 = -\frac{1}{3}$
解：对于方程$3x(2x + 1)=4x + 2，$变形为$3x(2x + 1)-2(2x + 1)=0，$提取公因式得$(2x + 1)(3x - 2)=0，$则$2x + 1 = 0$或$3x - 2 = 0，$解得$x_1 = -\frac{1}{2},x_2=\frac{2}{3}$
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