b) (x-1)(x+2)(x+3)(x+6)+20

=(x-1)(x+6) (x+2)(x+3)+20

=[x(x+6)-1(x+6)][x(x+3)+2(x+3)]+20

=(x²+6x-x-6)(x²+3x+2x+6)+20

=(x²+5x-6)(x²+5x+6)+20

Đặt x2+5x=a =>x²+5x-6=a-6 và x²+5x+6=a+6

ta có(a-6)(a+6)+20

=a²-6²+20

=a²-36+20

=a²-16=a²-4²

=(a-4)(a+4)

thay a=x²+5x vào ta đc

(x²+5x-4)(x²+5x+4)

xấu như ma ấy

2) x³-9x²+6x+16

=x³+x²-10x²-10x+16x+16

=(x³+x²)-(10x²+10x)+(16x+16)

=x²(x+1)-10x(x+1)+16(x+1)

=(x+1)(x²-10x+16)

=(x+1)(x²-2x-8x+16)

=(x+1)[(x(x-2)-8(x-2)]

=(x+1)(x-8)(x-2)

a)x³-7x+6

=x³+0x²-7x+6

=x³-x²+x²-x-6x+6

=(x³-x²)+(x²-x)-(6x-6)

=x²(x-1)+x(x-1)-6(x-1)

=(x-1)(x²+x-6)

=(x-1)(x²-2x+3x-6)

=(x-1)[x(x-2)+3(x-2)]

=(x-1)(x+3)(x-2)

0,6

2978,4

86,7

16x²-9(x+1)²=0

=>(4x)²-[3(x+1)]²=0

=>(4x)²-(3x+3)²=0

=>[4x-(3x+3)][4x+(3x+3)]=0

=>(4x-3x-3)(4x+3x+3)=0

=>(x-3)(7x+3)=0

=>x-3=0 hoặc 7x+3=0

=>x=3 hoặc 7x=-3=>x=-\(\dfrac{3}{7}\)

vậy x=3 hoặc x=-\(\dfrac{3}{7}\)