初二代数分解因式,例一到例四求解

2个回答

  • 9x^4-3x^3+7x^2-3x-2

    =9x^4-3x^3-2x^2+9x^2-3x-2

    =x^2(9x^2-3x-2)+(9x^2-3x-2)

    =(x^2-1)(9x^2-3x-2)

    =(x-1)(x+1)(9x^2-3x-2)

    x^2+3xy+2y^2+4x+5y+3

    =(x+y)(x+2y)+3(x+y)+(x+2y+3)

    =(x+y)(x+2y+3)+(x+2y+3)

    =(x+y+1)(x+2y+3)

    x^4-2x^3-27x^2-44x+7

    设原式=(x^2+ax+1)(x^2+bx+7)或=(x^2+ax-1)(x^2+bx-7)

    x^4+(a+b)x^3+(ab+8)x^2+(7a+b)x+7=x^4-2x^3-27x^2-44x+7

    或x^4+(a+b)x^3+(ab-8)x^2-(7a+b)x+7=x^4-2x^3-27x^2-44x+7

    可得方程组

    a+b=-2

    ab+8=-27

    7a+b=-44

    或a+b=-2

    ab-8=-27

    -7a-b=-44

    解第一个方程组可得a=-7 b=5,第二个无解

    所以原式=(x²-7x+1)(x²+5x+7)

    x+y+z=3a

    (x-a)+(y-a)+(z-a)=0

    设x-a=m,y-a=n,z-a=p

    则m+n+p=0 p=-(m+n)

    [(x-a)(y-a)+(y-a)(z-a)+(z-a)(x-a)]/[(x-a)^2+(y-a)^2+(z-a)^2]

    =(mn+np+pm)/(m^2+n^2+p^2)

    =[mn+p(m+n)]/(m^2+n^2+p^2)

    =[mn-(m+n)^2]/[m^2+n^2+(m+n)^2]

    =(mn-m^2-2mn-n^2)/2(m^2-mn+n^2)

    =-(m^2-mn+n^2)/2(m^2-mn+n^2)

    =-1/2