Giúp mình với! dạng hằng đẳng thức đáng nhớ

a) $(4x-1)^2-4(2x-3)^2-x-4=0$ $(4x-1)^2-[2(2x-3)]^2-x-4=0$ $(4x-1)^2-(4x-6)^2-x-4=0$ $(4x-1-4x+6)(4x-1+4x-6)-x-4=0$ $5(8x-7)-x-4=0$ $40x-35-x-4=0$ $39x=39$ $x=1$ b) $x(x-5)-(4-x)^2=7x+1$ $x(x-5)-(4-x)^2-(7x+1)=0$ $x(x-5)-(4-x)^2-(x+6x+1)=0$ $x(x-5)-(4-x)^2-(x+6(x+1))=0$ $x(x-5)-(4-x)^2-(x+(4-3+x)(x+1))=0$ $x(x-5)-(4-x)^2-(x+(4-x)(x+1)+3(x+1))=0$ $x(x-5)-(4-x)^2-(x+(4-x)(x+1)+(4-1)(x+1))=0$ $x(x-5)-(4-x)^2-(x+(4-x)(x+1)+(4-x)(x+1))=0$ $x(x-5)-(4-x)^2-(x+2(4-x)(x+1))=0$ $x(x-5)-(4-x)^2-(x+2(4-x)(x+1))=0$ $(x-5-(4-x))(x-(4-x))=0$ $(2x-9)(2x-4)=0$ $2x-9=0$ hoặc $2x-4=0$ $x=\frac{9}{2}$ hoặc $x=2$ c) $(2x-6)(x+3)=2(x-3)^2$ $(2(x-3))(x+3)=2(x-3)^2$ $(x-3)(x+3)=(x-3)^2$ $(x-3)(x+3-x+3)=0$ $(x-3)\times 6=0$ $x-3=0$ $x=3$