Задание 455

Решение:

• \({\largeа)}\ 16a^2bc^3-12ac^3+28b^2c^2-8abc^5=4c^2(4a^2bc-3ac+7b^2-2abc^3)\)

• \({\largeб)}\ 12x^2yz+18xy^3z^2-27x^5z^6-24xy^4z^4=3xz(4xy+6y^3z-9x^4z^5-8y^4z^3)\)

• \({\largeв)}\ 0{,}25m^2n^2k-0{,}45m^3nk^2-1{,}5mn^3k^2-0{,}05m^5n^3k=0{,}05mnk(5mn-9m^2k-30n^2k-m^4n^2)\)

• \({\largeг)}\ 1{,}42x^2y^4z^3-2\frac{1}{2}xy^3z^2-0{,}2x^3y^2z+3\frac{1}{3}xy^3z^2=1{,}42x^2y^4z^3-\frac{5}{2}xy^3z^2-0{,}2x^3y^2z+\frac{10}{3}xy^3z^2=0{,}2xy^2z\left(7{,}1xy^2z^2-\frac{25}{2}yz-x^2+\frac{50}{3}yz\right)=0{,}2xy^2z\left(7{,}1xy^2z^2-12\frac{1}{2}yz-x^2+16\frac{2}{3}yz\right)\)

• \({\largeд)}\ \frac{1}{3}a^2bx^3-1\frac{1}{2}ab^2x^2+0{,}3a^2x^3-1{,}1a^5b^3x^4=\frac{1}{3}a^2bx^3-\frac{3}{2}ab^2x^2+0{,}3a^2x^3-1{,}1a^5b^3x^4=\frac{1}{3}ax^2\left(abx-\frac{9}{2}b^2+0{,}9ax-3{,}3a^4b^3x^2\right)=\frac{1}{3}ax^2\left(abx-4\frac{1}{2}b^2+0{,}9ax-3{,}3a^4b^3x^2\right)\)