1. Алдымен жақшадағы өрнекті стандарт түрдегі көпмүшеге түрлендіреміз:

\(\displaystyle \begin{aligned}3y^{\,24}-y^{\,5}\cdot 5y^{\, 8}-2y^{\,2}+4y&=3y^{\,24}-5\cdot (\,y^{\,5}\cdot y^{\, 8})-2y^{\,2}+4y=\\&=3y^{\,24}-5\cdot y^{\,5+8}-2y^{\,2}+4y=\\&=3y^{\,24}-5y^{\,13}-2y^{\,2}+4y{\small .}\end{aligned}\)

Сонда

\(\displaystyle 10y^{\,2}\cdot (3y^{\,24}-y^{\,5}\cdot 5y^{\, 8}-2y^{\,2}+4y\,)\cdot 7y^{\,3}=10y^{\,2}\cdot (3y^{\,24}-5y^{\,13}-2y^{\,2}+4y\,)\cdot 7y^{\,3}{\small .}\)

2. Енді жақшаларды ашайық.

\(\displaystyle 10y^{\,2}{\small }\) көбейтіп, нәтижені стандарт түрге келтіреміз:

\(\displaystyle \begin{array}{l}\color{blue}{10y^{\,2}}\cdot (3y^{\,24}-5y^{\,13}-2y^{\,2}+4y\,)\cdot 7y^{\,3}=\\\kern{6em} =(\color{blue}{10y^{\,2}}\cdot 3y^{\,24}-\color{blue}{10y^{\,2}}\cdot 5y^{\,13}-\color{blue}{10y^{\,2}}\cdot 2y^{\,2}+\color{blue}{10y^{\,2}}\cdot 4y\,)\cdot 7y^{\,3}=\\\kern{6em} =((10\cdot 3)\cdot (\,y^{\,2}\cdot y^{\,24})-(10\cdot 5)\cdot (\,y^{\,2}\cdot y^{\,13})-\\\kern{15em} -(10\cdot 2)\cdot (\,y^{\,2}\cdot y^{\,2})+(10\cdot 4)\cdot (\,y^{\,2}\cdot y\,))\cdot 7y^{\,3}=\\\kern{6em} =(30\cdot y^{\,2+24}-50\cdot y^{\,2+13}-20\cdot y^{\,2+2}+40\cdot y^{\,2+1})\cdot 7y^{\,3}=\\\kern{18em} =(30y^{\,26}-50y^{\,15}-20y^{\,4}+40y^{\,3})\cdot 7y^{\,3}{\small ;}\end{array}\)

\(\displaystyle 7y^{\,3}{\small }\) көбейтіп, нәтижені стандарт түрге келтіреміз:

\(\displaystyle \begin{array}{l}(30y^{\,26}-50y^{\,15}-20y^{\,4}+40y^{\,3})\cdot \color{green}{7y^{\,3}}=\\\kern{6em} =30y^{\,26}\cdot \color{green}{7y^{\,3}}-50y^{\,15}\cdot \color{green}{7y^{\,3}}-20y^{\,4}\cdot \color{green}{7y^{\,3}}+40y^{\,3}\cdot \color{green}{7y^{\,3}}=\\\kern{6em} =(30\cdot 7)\cdot (\,y^{\,26}\cdot y^{\,3})-(50\cdot 7)\cdot (\,y^{\,15}\cdot y^{\,3})-\\\kern{15em} -(20\cdot 7)\cdot (\,y^{\,4}\cdot y^{\,3})+(40\cdot 7)\cdot (\,y^{\,3}\cdot y^{\,3})=\\\kern{6em} =210\cdot y^{\,26+3}-350\cdot y^{\,15+3}-140\cdot y^{\,4+3}+280\cdot y^{\,3+3}=\\\kern{18em} =210y^{\,29}-350y^{\,18}-140y^{\,7}+280y^{\,6}{\small .}\end{array}\)

Осылайша,

\(\displaystyle 10y^{\,2}\cdot (3y^{\,24}-y^{\,5}\cdot 5y^{\, 8}-2y^{\,2}+4y\,)\cdot 7y^{\,3}=210y^{\,29}-350y^{\,18}-140y^{\,7}+280y^{\,6}{\small .}\)

Жауабы: \(\displaystyle 210y^{\,29}-350y^{\,18}-140y^{\,7}+280y^{\,6}{\small .}\)