1. Әрбір қосылғышты \(\displaystyle 3xy^{\,2}\) көбейте отырып және алынған әрбір бірмүшені стандарт түрге түрлендіре отырып, ішкі жақшаларды ашайық:

\(\displaystyle \begin{array}{l}\color{blue}{3xy^{\,2}}(2xz^{\,2}-yz^{\,3}+4xyz\,)=\color{blue}{3xy^{\,2}}\cdot 2xz^{\,2}-\color{blue}{3xy^{\,2}}\cdot yz^{\,3}+\color{blue}{3xy^{\,2}}\cdot 4xyz=\\\;\phantom{\color{blue}{3xy^{\,2}}(2xz^{\,2}-yz^{\,3}+4xyz\,)}=(3\cdot 2)\cdot (x\cdot x\,)\cdot y^{\,2}z^{\,2}-3\cdot x\cdot (\,y^{\,2}\cdot y\,)\cdot z^{\,3}+\\\kern{25em} +(3\cdot 4)\cdot (x\cdot x\,)\cdot (\,y^{\,2}\cdot y\,)\cdot z=\\\;\phantom{\color{blue}{3xy^{\,2}}(2xz^{\,2}-yz^{\,3}+4xyz\,)}=6\cdot x^{\,1+1}\cdot y^{\,2}z^{\,2}-3\cdot x\cdot y^{\,2+1}\cdot z^{\,3}+12\cdot x^{\,1+1}\cdot y^{\,2+1}\cdot z=\\\;\phantom{\color{blue}{3xy^{\,2}}(2xz^{\,2}-yz^{\,3}+4xyz\,)}=6x^{\,2}y^{\,2}z^{\,2}-3xy^{\,3}z^{\,3}+12x^{\,2}y^{\,3}z{\small .}\end{array}\)

Сонда

\(\displaystyle \begin{array}{l}-5x^{\,2}yz^{\,2}(11xyz^{\,2}+3xy^{\,2}(2xz^{\,2}-yz^{\,3}+4xyz\,))=\\\kern{15em} =-5x^{\,2}yz^{\,2}(11xyz^{\,2}+6x^{\,2}y^{\,2}z^{\,2}-3xy^{\,3}z^{\,3}+12x^{\,2}y^{\,3}z\,){\small .}\end{array}\)

2. Тағы да жақшадағы әрбір қосылғышты  \(\displaystyle -5x^{\,2}yz^{\,2}{\small }\) көбейту арқылы жақшаларды ашайық:

\(\displaystyle \begin{array}{l}\color{blue}{ -5x^{\,2}yz^{\,2}}(11xyz^{\,2}+6x^{\,2}y^{\,2}z^{\,2}-3xy^{\,3}z^{\,3}+12x^{\,2}y^{\,3}z\,)=\\\kern{3em} =(\color{blue}{-5x^{\,2}yz^{\,2}})\cdot 11xyz^{\,2}+(\color{blue}{-5x^{\,2}yz^{\,2}})\cdot 6x^{\,2}y^{\,2}z^{\,2}-\\\kern{18em} -(\color{blue}{-5x^{\,2}yz^{\,2}})\cdot 3xy^{\,3}z^{\,3}+(\color{blue}{-5x^{\,2}yz^{\,2}})\cdot 12x^{\,2}y^{\,3}z=\\\kern{3em} =((-5)\cdot 11)\cdot (x^{\,2}\cdot x\,)\cdot (\,y\cdot y\,)\cdot (z^{\,2}\cdot z^{\,2})+((-5)\cdot 6)\cdot (x^{\,2}\cdot x^{\,2})\cdot (\,y\cdot y^{\,2})\cdot (z^{\,2}\cdot z^{\,2})-\\\kern{4em} -((-5)\cdot 3)\cdot (x^{\,2}\cdot x\,)\cdot (\,y\cdot y^{\,3})\cdot (z^{\,2}\cdot z^{\,3})+((-5)\cdot 12)\cdot (x^{\,2}\cdot x^{\,2})\cdot (\,y\cdot y^{\,3})\cdot (z^{\,2}\cdot z\,)=\\\kern{3em} =-55\cdot x^{\,2+1}\cdot y^{\,1+1}\cdot z^{\,2+2}-30\cdot x^{\,2+2}\cdot y^{\,1+2}\cdot z^{\,2+2}+\\\kern{15em} +15\cdot x^{\,2+1}\cdot y^{\,1+3}\cdot z^{\,2+3}-60\cdot x^{\,2+2}\cdot y^{\,1+3}\cdot z^{\,2+1}=\\\kern{16em} =-55x^{\,3}y^{\,2}z^{\,4}-30x^{\,4}y^{\,3}z^{\,4}+15x^{\,3}y^{\,4}z^{\,5}-60x^{\,4}y^{\,4}z^{\,3}{\small .}\end{array}\)

Осылайша,

\(\displaystyle \begin{array}{l}-5x^{\,2}yz^{\,2}(11xyz^{\,2}+3xy^{\,2}(2xz^{\,2}-yz^{\,3}+4xyz\,))=\\\kern{15em} =-55x^{\,3}y^{\,2}z^{\,4}-30x^{\,4}y^{\,3}z^{\,4}+15x^{\,3}y^{\,4}z^{\,5}-60x^{\,4}y^{\,4}z^{\,3}{\small .}\end{array}\)

Жауабы: \(\displaystyle -55x^{\,3}y^{\,2}z^{\,4}-30x^{\,4}y^{\,3}z^{\,4}+15x^{\,3}y^{\,4}z^{\,5}-60x^{\,4}y^{\,4}z^{\,3}{\small .}\)