а4) = ba-bc-3a+3c=a(b-3)-c(b-3)=(b-3)(a-c)
a5)=а(х+у)-b(x+y)=(a-b)(x+y)
с1)=х²(а+b)-y²(a+b)-y(a+b)=(a+b)(x²-y²-y)
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а4) = ba-bc-3a+3c=a(b-3)-c(b-3)=(b-3)(a-c)
a5)=а(х+у)-b(x+y)=(a-b)(x+y)
с1)=х²(а+b)-y²(a+b)-y(a+b)=(a+b)(x²-y²-y)