Задание 316

Решение:

• \({\largeа)}\ (5ab^2+4b^3)(3ab^3-4a^2)-18a^2b^3=15a^2b^5+12ab^6-20a^3b^2\ -\) \(16a^2b^3\) \(-\) \(18a^2b^3\) \(=15a^2b^5+12ab^6-20a^3b^2-34a^2b^3\)

• \({\largeб)}\ (7x^3y^2-xy)(-2x^2y^2+5xy^3)+12x^5y^4={-}\) \(14x^5y^4\) \(+\ 2x^3y^3+35x^4y^5-5x^2y^4\ +\) \(12x^5y^4\) \(={-}2x^5y^4+2x^3y^3+35x^4y^5-5x^2y^4\)

• \({\largeв)}\ (x^3+x^2y+xy^2+y^3)(x-y)-x^2y(x-y)=(x-y)(x^3+x^2y+xy^2+y^3-x^2y)=(x-y)(x^3+xy^2+y^3)=x^4-x^3y+x^2y^2\ -\) \(xy^3\) \(+\) \(xy^3\) \(-\ y^4=x^4-x^3y+x^2y^2-y^4\)

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

• \({\largeд)}\ 2-(-4x+1)(x-1)+2(6x-4)(x+3)=2-(-4x^2+x+4x-1)+2(6x^2-4x+18x-12)=2-(-4x^2+5x-1)+2(6x^2+14x-12)=\) \(2\) \(+\) \(4x^2\) \(-\) \(5x\) \(+\) \(1\) \(+\) \(12x^2\) \(+\) \(28x\) \(-\) \(24\) \(=16x^2+23x-21\)

• \({\largeе)}\ 6(x+1)(x+1)+2(x-1)(x^2+x+1)-2(x+1)=6(x^2+x+x+1)+2(x^3-x^2+x^2-x+x-1)-2x-2=6(x^2+2x+1)+2(x^3-1)-2x-2=6x^2\ +\) \(12x\) \(+\) \(6\) \(+\ 2x^3\ -\) \(2\) \(-\) \(2x\) \(-\) \(2\) \(=2x^3+6x^2+10x+2\)

• \({\largeж)}\ (x+2)(x^2-2x+4)-x(x-3)(x+3)=(x^3+2x^2-2x^2-4x+4x+8)-x(x^2-3x+3x-9)=(x^3+8)-x(x^2-9)=\) \(x^3\) \(+\ 8\ -\) \(x^3\) \(+\ 9x=9x+8\)

• \({\largeз)}\ 3(3x-1)(2x+5)-6(2x-1)(x+2)=3(6x^2-2x+15x-5)-6(2x^2-x+4x-2)=3(6x^2+13x-5)-6(2x^2+3x-2)=\) \(18x^2\) \(+\) \(39x\) \(-\) \(15\) \(-\) \(12x^2\) \(-\) \(18x\) \(+\) \(12\) \(=6x^2+21x-3\)

• \({\largeи)}\ (x^2+2)(x^2+2)-(x-2)(x+2)(x^2+4)=(x^4+2x^2+2x^2+4)-(x^2-2x+2x-4)(x^2+4)=(x^4+4x^2+4)-(x^2-4)(x^2+4)=x^4+4x^2+4-(x^4-4x^2+4x^2-16)=x^4+4x^2+4-(x^4-16)=\) \(x^4\) \(+\ 4x^2\ +\) \(4\) \(-\) \(x^4\) \(+\) \(16\) \(=4x^2+20\)

• \({\largeк)}\ 5(a-2)(a+2)-\frac{1}{2}(8a-6)(8a-6)+17=5(a^2-2a+2a-4)-\frac{1}{2}(64a^2-48a-48a+36)+17=5(a^2-4)-\frac{1}{2}(64a^2-96a+36)+17=\) \(5a^2\) \(-\) \(20\) \(-\) \(32a^2\) \(+\ 48a\ -\) \(18\) \(+\) \(17\) \(={-}27a^2+48a-21\)