Question

因式分解的数学题

1.ab-a+b-1
2.3m+nx-3n-mx
3.xy-xz-y^2+2yz-z^2

Answer 1

1.ab-a+b-1
=a(b-1)+(b-1)
=(a+1)(b-1)

2.3m+nx-3n-mx
=(3m-3n)+nx-mx
=3(m-n)-x(m-n)
=(3-x)(m-n)

3.xy-xz-y²+2yz-z²
=x(y-z)-(y-z)²
=[x-(y-z)](y-z)
=(x-y+z)(y-z)

Answer 2

1.ab-a+b-1
=(a+1)(b-1)
2.3m+nx-3n-mx
=(3-x)(m-n)
3.xy-xz-y^2+2yz-z^2
=x(y-z)-(y-z)^2
=(y-z)(x-y+z)

Answer 3

1.ab-a+b-1=a(b-1)+(b-1)=(a+1)(b-1)
2.3m+nx-3n-mx=3(m-n)-x(m-n)=(m-n)(3-x)
3.xy-xz-y^2+2yz-z^2=x(y-z)-(y-z)^2=(x-y+z)(y-z)

Answer 4

1.
原式=a(b-1)+(b-1)=(a+1)(b-1)
2.
原式=3(m-n)-(m-n)x=(3-x)(m-n)
3.
原式=x(y-z)-(y-z)²=(x-y+z)(y-z)

Answer 5

1.ab-a+b-1=a(b-1)+（b-1)=(a+1)(b-1)
2.3m+nx-3n-mx=m(3-x)-n(3-x)=(m-n)(3-x)
3.xy-xz-y^2+2yz-z^2=-(y-z-x)（y-z）

Answer 6

1.原式=ab+b-(a+1)=b(a+1)-(a+1)=(a+1)(b-1)
2.原式=3m-3n+nx-mx=3(m-n)-x(m-n)=(m-n)(3-x)
3.原式=x(y-z)-(y^2-2yz+z^2)=x(y-z)-(y-z)^2=x(y-z)(x-y+z)