(x²+6x+8)(x²+8x+15)-24

<=>(x²+4x+2x+8)(x²+5x+3x+15)-24

<=> [x(x+4)+2(x+4)][x(x+5)+3(x+5)]-24

<=> (x+4)(x+2)(x+5)(x+3)-24

<=> (x+4)(x+3)(x+2)(x+5)-24

<=>(x²+7x+12)(x²+7x+10)

đặt t=x²+7x+11 ta có:

(t-1)(t+1)-24

<=> t²-1-24

<=>t²-25

<=>(t-5)(t+5)

thay t=x²+7x+11 vào ta có:

(x²+7x+11-5)(x²+7x+11+5)

<=>(x²+7x+6)(x²+7x+16)

(x-1)(x+2)(x+3)(x+6)-28

<=> (x-1)(x+6)(x+2)(x+3)-28

<=> (x²+5x-6)(x²​+5x+6)-28

đặt x²​+5x=t ta có:

<=>(t-6)(t+6)-28

<=>t²-36-28

<=>t²-64

<=>(t-8)(t+8)

thay t=x²+5x và đa thức ta có:

(x²​+5x-8)(x²​+5x+8)

ta có: x²+y²+xy+3x-3y+9=0

=> 2(x²+y²+xy+3x-3y+9)=0.2

=>2x²+2y²+2xy+6x-6y+18=0

<=>(x²+xy+y²)+(x²+6x+9)+(y²-6x+9)=0

<=>(x+y)²+(x+3)²+(y-3)²=0

=> (x+y)²=(x+3)²=(y-3)²=0

=> x= -3,y=3

thay vào A ta có:

A=(-3+3+1)²⁰¹⁷+(-3+2)²⁰¹⁸

A=1²⁰¹⁷+(-1)²⁰¹⁹

A=1-1

A=0