= (x^2 -y^2)-2.(x-y)

=(x-y).(x+y)-2.(x-y)

=(x-y).(x+y-2)

a,=(x+2).(x^2-2x+2^2)-18-x^3

=x^3 + 2^3 - 18 -x^3=(x^3-x^3)+(8-18) = -10

b, =(2x-y).((2x)²+2xy +y²) - (2x +y).((2x)^2-2xy +y^2)

=(2x)³-y³- (2x)³-y³= -2 .y³

2 ý này áp dụng HĐT : x³+y³=(x+y).(x²-xy+y²)

x³-y³=(x-y).(x²+xy+y²)

b, = x^2 +4x +5x+20

=x.(x+4)+5.(x+4)

=(x+5).(x+4)

a, = (2y)² - (x^2 - 2x +1)

= (2y)² - (x-1)²

=(2y - x +1).(2y + x - 1)

đặt a=x^2-3x+5 , b= x^2 -3x-1 (1)

sau đó thay vào sẽ được biểu thức : a² -2ab+b² =(a-b)² (2)

^{sau đó thay (1) vào (2) , tiếp đó rút gon bt vùa thay sẽ suy ra đpcm}

= (x²)² +2.2x² +2² - 4x²

^{= (x^2 +2)^2 -(2x)^2}

^{=(x^2 +2 -2x).(x^2+2+2x)}

a, = x³ - x² + x² - x + 5x - 5

= x².(x-1) +x.(x-1)+5.(x-1)

= (x-1).(x² + x +5)

b,=x.(x² - 6x +9)

= x.(x² -2.3.x +3²)=x.(x-3)²

^{c, = (x^2 -y^2 ) -(x-y) = (x-y).(x+y)-(x-y)=(x-y).(x+y-1)}

a, 4xⁿy³. (2x² - 3y²)

b, (a-b-x+3).(a-b+x-3)

c, (3x-5-x+3)(3x-5+x-3)=(2x-2)(4x-8)=8.(x-1)(x-2)

= 6x^2 - 9x - 2x + 3

= 3x(2x-3) -(2x-3)

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

Ta có : a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd)