Bài 1:
a)x3-19x-30
=x3+5x2+6x-5x2-25x-30
=x(x2+5x+6)-5(x2+5x+6)
=(x-5)(x2+5x+6)
=(x-5)(x2+3x+2x+6)
=(x-5)[x(x+3)+2(x+3)]
=(x-5)(x+2)(x+3)
b)x4+x2+1
=(x2)2-2x2+2x2+1+x2
=(x2+1)2-2x2+x2
=(x2+1)2-x2
=[(x2+1)+x]*[(x2+1)-x]
=(x2+x+1)(x2-x+1)
Bài 2:
2x2=x
=>2x2-x=0
=>x(2x-1)=0
=>x=0 hoặc 2x-1=0
=>x=0 hoặc \(x=\frac{1}{2}\)