1. Алдымен барлық көбейткіштер мен қосылғыштарды стандарт түрдегі бірмүшелерге түрлендіреміз:

• \(\displaystyle yz^{\,2}\cdot xz^{\,3}=x\cdot y\cdot (z^{\,2}\cdot z^{\,3})=x\cdot y\cdot z^{\,2+3}=xyz^{\,5}{\small ,}\)
• \(\displaystyle 3x^{\,2}z^{\,2}\cdot xy\cdot xz^{\,3}=3\cdot (x^{\,2}\cdot x\cdot x\,)\cdot y\cdot (z^{\,2}\cdot z^{\,3})=3\cdot x^{\,2+1+1}\cdot y\cdot z^{\,2+3}=3x^{\,4}yz^{\,5}{\small .}\)

Сонда

2. Енді әр жақшаны олардың көбейткішіне көбейтейік.

Бірінші жақшаны \(\displaystyle xyz^{\,5}{\small }\) көбейтіп, нәтижені  стандарт түрге келтіреміз:

\(\displaystyle \begin{array}{l}\color{blue}{xyz^{\,5}}\cdot (3x^{\,3}y^{\,3}-4y^{\,2}z^{\,4}+5x^{\,2}y^{\,2}z^{\,4})=\\\kern{3em} =\color{blue}{xyz^{\,5}}\cdot 3x^{\,3}y^{\,3}-\color{blue}{xyz^{\,5}}\cdot 4y^{\,2}z^{\,4}+\color{blue}{xyz^{\,5}}\cdot 5x^{\,2}y^{\,2}z^{\,4}=\\\kern{3em} =3\cdot (x\cdot x^{\,3})\cdot (\,y\cdot y^{\,3})\cdot z^{\,5}-4\cdot x\cdot (\,y\cdot y^{\,2})\cdot (z^{\,5}\cdot z^{\,4})+5\cdot (x\cdot x^{\,2})\cdot (\,y\cdot y^{\,2})\cdot (z^{\,5}\cdot z^{\,4})=\\\kern{3em} =3\cdot x^{\,1+3}\cdot y^{\,1+3}\cdot z^{\,5}-4\cdot x\cdot y^{\,1+2}\cdot z^{\,5+4}+5\cdot x^{\,1+2}\cdot y^{\,1+2}\cdot z^{\,5+4}\cdot z^{\,4}=\\\kern{23em} =3x^{\,4}y^{\,4}z^{\,5}-4xy^{\,3}z^{\,9}+5x^{\,3}y^{\,3}z^{\,9}{\small .}\end{array}\)

Екінші жақшаны \(\displaystyle x^{\,2}z^{\,2}{\small ,}\) көбейтіп, нәтижені стандарт түрге келтіреміз:

\(\displaystyle \begin{array}{l}(2x^{\,2}y^{\,4}z^{\,3}+3x^{\,4}yz^{\,5}-2xy^{\,3}z^{\,7})\cdot \color{blue}{x^{\,2}z^{\,2}}=\\\kern{3em} =2x^{\,2}y^{\,4}z^{\,3}\cdot \color{blue}{x^{\,2}z^{\,2}}+3x^{\,4}yz^{\,5}\cdot \color{blue}{x^{\,2}z^{\,2}}-2xy^{\,3}z^{\,7}\cdot \color{blue}{x^{\,2}z^{\,2}}=\\\kern{3em} =2\cdot (x^{\,2}\cdot x^{\,2})\cdot y^{\,4}\cdot (z^{\,3}\cdot z^{\,2})+3\cdot (x^{\,4}\cdot x^{\,2})\cdot y\cdot (z^{\,5}\cdot z^{\,2})-2\cdot (x\cdot x^{\,2})\cdot y^{\,3}\cdot (z^{\,7}\cdot z^{\,2})=\\\kern{3em} =2\cdot x^{\,2+2}\cdot y^{\,4}\cdot z^{\,3+2}+3\cdot x^{\,4+2}\cdot y\cdot z^{\,5+2}-2\cdot x^{\,1+2}\cdot y^{\,3}\cdot z^{\,7+2}=\\\kern{23em} =2x^{\,4}y^{\,4}z^{\,5}+3x^{\,6}yz^{\,7}-2x^{\,3}y^{\,3}z^{\,9}{\small .}\end{array}\)

Сондықтан

\(\displaystyle \begin{array}{l}xyz^{\,5}(3x^{\,3}y^{\,3}-4y^{\,2}z^{\,4}+5x^{\,2}y^{\,2}z^{\,4})-(2x^{\,2}y^{\,4}z^{\,3}+3x^{\,4}yz^{\,5}-2xy^{\,3}z^{\,7})x^{\,2}z^{\,2}=\\\kern{6em} =(3x^{\,4}y^{\,4}z^{\,5}-4xy^{\,3}z^{\,9}+5x^{\,3}y^{\,3}z^{\,9})-(2x^{\,4}y^{\,4}z^{\,5}+3x^{\,6}yz^{\,7}-2x^{\,3}y^{\,3}z^{\,9}){\small .}\end{array}\)

3. Жақшаларды ашып, ұқсас қосылғыштарды келтіре отырып, алынған өрнекті ықшамдайық. Екінші жақшаның алдында минус таңбасы тұрғандықтан, осы жақшалардың ішіндегі барлық таңбалар қарама-қарсыға өзгереді:

\(\displaystyle \begin{array}{l}(3x^{\,4}y^{\,4}z^{\,5}-4xy^{\,3}z^{\,9}+5x^{\,3}y^{\,3}z^{\,9})-(2x^{\,4}y^{\,4}z^{\,5}+3x^{\,6}yz^{\,7}-2x^{\,3}y^{\,3}z^{\,9})=\\\kern{6em} =3\color{blue}{x^{\,4}y^{\,4}z^{\,5}}-4xy^{\,3}z^{\,9}+5\color{green}{x^{\,3}y^{\,3}z^{\,9}}-2\color{blue}{x^{\,4}y^{\,4}z^{\,5}}-3x^{\,6}yz^{\,7}+2\color{green}{x^{\,3}y^{\,3}z^{\,9}}=\\\kern{6em} =(3\color{blue}{x^{\,4}y^{\,4}z^{\,5}}-2\color{blue}{x^{\,4}y^{\,4}z^{\,5}})-4xy^{\,3}z^{\,9}+(5\color{green}{x^{\,3}y^{\,3}z^{\,9}}+2\color{green}{x^{\,3}y^{\,3}z^{\,9}})-3x^{\,6}yz^{\,7}=\\\kern{6em} =(3-2)\color{blue}{x^{\,4}y^{\,4}z^{\,5}}-4xy^{\,3}z^{\,9}+(5+2)\color{green}{x^{\,3}y^{\,3}z^{\,9}}-3x^{\,6}yz^{\,7}=\\\kern{17em} =\color{blue}{x^{\,4}y^{\,4}z^{\,5}}-4xy^{\,3}z^{\,9}+7\color{green}{x^{\,3}y^{\,3}z^{\,9}}-3x^{\,6}yz^{\,7}{\small .}\end{array}\)

Осылайша,

\(\displaystyle \begin{array}{l}yz^{\,2}\cdot xz^{\,3}(3x^{\,3}y^{\,3}-4y^{\,2}z^{\,4}+5x^{\,2}y^{\,2}z^{\,4})-(2x^{\,2}y^{\,4}z^{\,3}+3x^{\,2}z^{\,2}\cdot xy\cdot xz^{\,3}-2xy^{\,3}z^{\,7})x^{\,2}z^{\,2}=\\\kern{20em} =x^{\,4}y^{\,4}z^{\,5}-4xy^{\,3}z^{\,9}+7x^{\,3}y^{\,3}z^{\,9}-3x^{\,6}yz^{\,7}{\small .}\end{array}\)

Жауабы: \(\displaystyle x^{\,4}y^{\,4}z^{\,5}-4xy^{\,3}z^{\,9}+7x^{\,3}y^{\,3}z^{\,9}-3x^{\,6}yz^{\,7}{\small .}\)