View attachment 140059bày e vs mn, thực sự ngu toán đại
((
a)[tex](x-6)^2-y^2= (x-6-y)(x-6+y)[/tex]
b) [tex]2(x^2-9)=2(x-3)(x+3)[/tex]
a) x(x+3)-2(x+3)=0=> (x-2)(x+3)=0=> x=2 hoac x=-3
b)[tex](x+3)(x^2-3x+9)+(x+3)(x-9)=0=>(x+3)(x^2-2x)=0 => x=-3 hoac x= 0 hoac x=2[/tex]