\(Bài1:\\ a,\left(2x-3\right)^2-4x\left(x-3\right)=4x^2-12x+9-4x^2+12=9\)
Bài 2:
\(a,3x^2-12xy=3x\left(x-4y\right)\\ b,x^2+7x-2\left(x+7\right)=x\left(x+7\right)-2\left(x+7\right)=\left(x+7\right)\left(x-2\right)\\ c,8x^3-8x^2+2x=2x\left(4x^2-4x+1\right)=2x\left(2x-1\right)^2\\ d,x^2-y^2+12y-36=x^2-\left(y^2-12y+36\right)=x^2-\left(y-6\right)^2=\left(x-y+6\right)\left(x+y-6\right)\)
\(1,\\ a,=4x^2-12x+9-4x^2+12x=9\\ b,=\left(15x^3-30x^2+20x^2-40x+41x-82+80\right):\left(x-2\right)\\ =\left[15x^2\left(x-2\right)+20x\left(x-2\right)+41\left(x-2\right)+80\right]:\left(x-2\right)\\ =15x^2+20x+41\left(\text{dư }80\right)\\ 2,\\ a,=3x\left(x-4y\right)\\ b,=x\left(x+7\right)-2\left(x+7\right)=\left(x-2\right)\left(x+7\right)\\ c,=2x\left(4x^2-4x+1\right)=2x\left(2x-1\right)^2\\ d,=x^2-\left(y-6\right)^2=\left(x-y+6\right)\left(x+y-6\right)\\ 3,\\ a,=\dfrac{x\left(x-6\right)\left(x+6\right)}{x\left(x+6\right)}=x-6\\ b,=\dfrac{2x+4+3x-6-18+5x}{\left(x+2\right)\left(x-2\right)}=\dfrac{10\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{10}{x+2}\)
