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  • 在實(shí)數(shù)范圍內(nèi)因式分解:(1)x^2-8xy+12y^2(2)x^2-7(3)x^4-36(4)x^4-6x^2+5(5)a^2-4a+1

    在實(shí)數(shù)范圍內(nèi)因式分解:(1)x^2-8xy+12y^2(2)x^2-7(3)x^4-36(4)x^4-6x^2+5(5)a^2-4a+1
    (6)x^3-4(x-2x^2)(7)2x^2-8xy+5y^2(8)3x^2y^2-5xy-1(9)-4x^2-8xy+y^2(10)-y^2x-4x^2y+2x^3
    數(shù)學(xué)人氣:274 ℃時(shí)間:2020-03-27 01:45:06
    優(yōu)質(zhì)解答
    x^2-8xy+12y^2=(x-6y)(x-2Y)
    x^2-7=(X+√7)(X-√7)
    x^4-36=(X^2+6)(x+√6)(x-√6)
    x^4-6x^2+5=(X+√5)(x-√5)(x+1)(x-1)
    a^2-4a+1=a^2-4a+1+3-3=(a-2)(a-2)-3=(a-2-√3)(a-2+√3)
    x^3-4(x-2x^2)=x^3+8x^2-4x=x(x^2+8x-4)=x(x^2+8x-4+16-16)=x(x+4-2√5)(x+4+2√5)
    2x^2-8xy+5y^2=2x^2-8xy+5y^2+x^2-x^2=(3x-5y)(x-y)-x^2=(√(3x-5y)(x-y)+x)(√(3x-5y)(x-y)-x)
    3x^2y^2-5xy-1=3x^2y^2-5xy-1+2x^2y^2-2x^2y^2=(√5xy-1)(√5xy+1)-2x^2y^2=[√(√5xy-1)(√5xy+1)+√2xy][√(√5xy-1)(√5xy+1)-√2xy]
    -4x^2-8xy+y^2=-4x^2-8xy+y^2+16x^2-16x^2=[√(4x+y)(4x-y)+2√5x][√(4x+y)(4x-y)-2√5x]
    -y^2x-4x^2y+2x^3=x(2x^2-4xy-y^2)=x[(√2x+y)(√2x-y)-4xy]
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