Question

Заключите два последних слагаемых в скобки, поставив перед ними знак "-", и затем выполните разложение на множители :

х(у+z)-2y-2z=

a(b+c)-b-c=

a(b-c)-4b+4c=

a(a-b)-ac+bc=

x(y-z)-y+z=

2b(x-y)+y-x=

5(c-b)+ab-ac=

2(x-c)-bx+bc=

Answer 1

x(y+z)-(2y+2z)

a(b+c)-(b+c)

a(b-c)-(4b-4c)

a(a-b)-(ac-bc)

x(y-z)-(y-z)

2b(x-y)-(-y+x)

5(c-d)-(-ab+ac)

2(x-c)-(bx-bc) 

Answer 2

1) х(у+z)-2y-2z=x(y+z)-(2y+2z)=x(y+z)-2(y+z)=(y+z)(x-2).

2) a(b+c)-b-c=a(b+c)-(b+c)=(b+c)(a-1).

3) a(b-c)-4b+4c=a(b-c)-(4b-4c)=a(b-c)-4(b-c)=(b-c)(a-4).

4) a(a-b)-ac+bc=a(a-b)-(ac-bc)=a(a-b)-c(a-b)=(a-b)(a-c).

5) x(y-z)-y+z=x(y-z)-(y-z)=(y-z)(x-1).

6) 2b(x-y)+y-x=2b(x-y)-(x-y)=(x-y)(2b-1).

7) 5(c-b)+ab-ac=5(c-b)-(ac-ab)=5(c-b)-a(c-b)=(c-b)(5-a).

8) 2(x-c)-bx+bc=2(x-c)-(bx-bc)=2(x-c)-b(x-c)=(x-c)(2-b).