Question

разложите на множители многочлены​

Answer 1

Ответ:

Объяснение:

$\displaystyle 1)\,\,16a^2b^3-32ab^2+64abc=(16ab)(ab^2)-(16ab)(2b)+(16ab)(4c)=\\=(16ab)(ab^2-2b+4c)\\\\2)\,\,9x^3y^4-27x^2yz+54xy^2=(9xy)(x^2y^3)-(9xy)(3xz)+(9xy)(6y)=\\=(9xy)(x^2y^3-3xz+6y)\\\\3)\,\,\frac3{16}a^4bc+\frac7{32}a^3bc-\frac9{64}a^4b^2c=(\frac1{64}a^3bc)(12a)+(\frac1{64}a^3bc)(14)-(\frac1{64}a^3bc)(9ab)=\\=(\frac1{64}a^3bc)(12a+14-9ab)\\\\4)\,\,\frac59x^8y^2z^3+\frac{11}{27}x^6y^3z^2-\frac{11}{54}x^7y^4z=(\frac{1}{54}x^6y^2z)(30x^2z^2)+(\frac{1}{54}x^6y^2z)(22yz)-(\frac{1}{54}x^6y^2z)(11xy^2)=$

$\displaystyle =(\frac{1}{54}x^6y^2z)(30x^2z^2+22yz-11xy^2)$