Question

分解因式:xy(x²-y²)+yz(y²-z²）+zx（z²-x²） 分解因式：4（x²+11x+30

分解因式:xy(x²-y²)+yz(y²-z²）+zx（z²-x²）
分解因式：4（x²+11x+30）（x²+22x+120）-3x²
求使x²-5x-24成为完全平方数的所有整数x的值
在线等！急！

Answer 1

xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=x^3y-xy^3+y^3z-yz^3+z^3x-zx^3
=x^3(y-z)+y^3(z-x)+z^3(x-y)
因为x-y=(x-z)+(z-y),所以：
=x^3(y-z)+y^3(z-x)+z^3[(x-z)+(z-y)]
=(x^3-z^3)(y-z)+(y^3-z^3)(z-x)
=(x-z)(x^2+xz+z^2)(y-z)+(y-z)(y^2+yz+z^2)(z-x)
=(x-z)(y-z)(x^2+zx+z^2-y^2-yz-z^2)
=(x-z)(y-z)(x-y)(x+y+z)

4(x^2+11x+30)(x^2+22x+120)-3x^2
=4(x^2+11x+30)(x^2+11x+30+11x+90)-3x^2

=4(x+5)(x+6)(x+10)(x+12)-3x^2
=4(x+5)(x+12)(x+6)(x+10)-3x^2
=4(x^2+17x+60)(x^2+16x+60)-3x^2
=4(x^2+16x+60+x)(x^2+16x+60)-3x^2
=4(x^2+16x+60)^2+x(x^2+16x+60)-3x^2
=[4(x^2+16x+60)-3x][(x^2+16x+60)+x]
=[4x^2+64x+240-3x][x^2+16x+60+x]
=(4x^2+61x+240)(x^2+17x+60)
=(4x^2+61x+240)(x+12)(x+5)