Задание 468

Решение:

• \({\largeа)}\ (2m+n)(6m+2n)-(m-3n)(8n+16m)=2(2m+n)(3m+n)-8(m-3n)(2m+n)=2(2m+n)(3m+n-4(m-3n))=2(2m+n)(3m+n-4m+12n)=2(2m+n)(13n-m)\)

• \({\largeб)}\ (x-1)(4x-6y)+(x+1)(18y-12x)=2(x-1)(2x-3y)+6(x+1)(3y-2x)=2(x-1)(2x-3y)-6(x+1)(2x-3y)=2(2x-3y)(x-1-3(x+1))=2(2x-3y)(x-1-3x-3)=2(2x-3y)({-}2x-4)=2(2x-3y)({-}2)(x+2)={-}4(2x-3y)(x+2)\)

• \({\largeв)}\ (2a+1)(5a-15)+(30-10a)(a-2)=5(2a+1)(a-3)+10(3-a)(a-2)=5(2a+1)(a-3)-10(a-3)(a-2)=5(a-3)(2a+1-2(a-2))=5(a-3)(2a+1-2a+4)=5(a-3)\cdot5=25(a-3)\)

• \({\largeг)}\ 2a(a+2)^2-3b(a+2)=(a+2)(2a(a+2)-3b)=(a+2)(2a^2+4a-3b)\)

• \({\largeд)}\ (x-2)^2(x-3)+(x-2)(x-3)^2=(x-2)(x-3)(x-2+x-3)=(x-2)(x-3)(2x-5)\)

• \({\largeе)}\ 3m(m+2n)-2n(m+2n)^2=(m+2n)(3m-2n(m+2n))=(m+2n)(3m-2mn-4n^2)\)

• \({\largeж)}\ (p+3q)^2(p-q)-(p+3q)(p-q)^2=(p+3q)(p-q)(p+3q-(p-q))=(p+3q)(p-q)(p+3q-p+q)=4q(p+3q)(p-q)\)