Заключите два последних слагаемых в скобки,поставив перед ними знак -,и затем выполните разложение на множители:x(y+z)-2y-3x ;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

Предыдущий
вопрос Следующий
вопрос

x(y+z)-2y-3x = x(y+z) - (2y+3x)
= xy + xz - 2y - 3x
= x(y+z) - 3(y+x)
= x(y+z) - 3(x+y)
= (y+z)x - 3(x+y)
= x(y+z-3)

a(b+c)-b-c = a(b+c) - (b+c)
= ab + ac - b - c
= a(b+c) - (b+c)
= b(a+c) - (b+c)
= (a+c)b - (b+c)
= b(a+c-1)

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

a(a-b)-ac+bc = a(a-b) - (ac-bc)
= a^2 - ab - ac + bc
= a(a-b-c) + bc
= a(a-b) - c(a-b)
= (a-c)(a-b)

x(y-z)-y+z = x(y-z) - (y-z)
= xy - xz - y + z
= x(y-z) - (y-z)
= -1(y-z)(x-1)
= -(z-y)(x-1)

2b(x-y)+y-x = 2b(x-y) - (y-x)
= 2bx - 2by - y + x
= -1(2by - 2bx + x - y)
= -(2b - 1)(y - x)

5(c-b)+ab-ac = 5(c-b) + (ab-ac)
= 5c - 5b + ab - ac
= -1(-5b + 5c - ac + ab)
= -(5b - ac + 5c - ab)

2(x-c)-bx+bc = 2(x-c) - (bx - bc)
= 2x - 2c - bx + bc
= -1(-2c + bc - bx + 2x)
= -(2c - bc + bx - 2x)