因式分解(x-1)^2-(2x+5)^2

第一道题用十字相乘法,第二道题用提公因式法
(x-1)²-(2x+5)²
=x²-2x+1-4x²-20x-25
=-(3x²+22x+24)
=-(x+6)(3x+4)

（ax+by)²+(bx-ay)²
=a²x²+2axby+b²y²+b²x²-2axby+a²y²
=a²x²+b²y²+b²x²+a²y²
=a²(x²+y²)+b²(x²+y²)
=(x²+y²)(a²+b²)

(x-1)^2-(2x+5)^2
=[(x-1)-(2x+5)][(x-1)+(2x+5)]
=(x-1-2x-5)(x-1+2x+5)

=(-x-6)(3x+4)
=-(x+6)(3x+4)

(ax+by)^2+(bx-ay)^2
=(ax)^2+2abxy+(by)^2+(bx)^2-2abxy+(ay)^2
=(ax)^2+(by)^2+(bx)^2+(ay)^2
=(ax)^2+(ay)^2+(by)^2+(bx)^2
=a^2x^2+a^2y^2+by^2+b^2x^2
=a^2(x^2+y^2)+b^2(x^2+y^2)
=(a^2+b^2)(x^2+y^2)

(x-1)^2-(2x+5)^2
=(x-1+2x+5)(x-1-2x-5)
=(3x+4)(-x-6)
=-(x+6)(3x+4)

（ax+by)^2+(bx-ay)^2
=a^2x^2+b^2y^2+b^2x^2+a^2y^2
=x^2(a^2+b^2)+y^2(a^2+b^2)
=(x^2+y^2)(a^2+b^2)

解：(x-1)^2-(2x+5)^2
=x^2-2x+1-4x^2-20x-25
=-3x^2-22x-24
=-(3x*x+18x+4x+4*6)
=-(3x+4)(x+6)

解：(ax+by)^2+(bx-ay)^2
=(ax)^2+2abxy+(by)^2+(bx)^2-2abxy+(ay)^2
=(a^2+b^2)x^2+(a^2+b^2)y^2
=(a^2+b^2)(x^2+y^2)

用公式法即可