Жақшадағы әрбір қосылғышты \(\displaystyle 5x^{\,2}y^{\,2}{\small }\) көбейту арқылы жақшаларды ашайық:

\(\displaystyle \begin{array}{l}\color{blue}{5x^{\,2}y^{\,2}}\cdot (6x^{\,3}y^{\,5}z+3xz^{\, 4}+xyz-11xz^{\,3})=\\\kern{9em} =\color{blue}{5x^{\,2}y^{\,2}}\cdot 6x^{\,3}y^{\,5}z+\color{blue}{5x^{\,2}y^{\,2}}\cdot 3xz^{\, 4}+\color{blue}{5x^{\,2}y^{\,2}}\cdot xyz-\color{blue}{5x^{\,2}y^{\,2}}\cdot 11xz^{\,3}{\small .}\end{array}\)

Қосылғыштарды стандарт түрдегі бірмүшелерге түрлендіру арқылы алынған өрнекті ықшамдайық:

\(\displaystyle \begin{array}{l}5x^{\,2}y^{\,2}\cdot 6x^{\,3}y^{\,5}z+5x^{\,2}y^{\,2}\cdot 3xz^{\, 4}+5x^{\,2}y^{\,2}\cdot xyz-5x^{\,2}y^{\,2}\cdot 11xz^{\,3}=\\\kern{5em} =(5\cdot 6)\cdot (x^{\,2}\cdot x^{\,3})\cdot (\,y^{\,2}\cdot y^{\,5})\cdot z+(5\cdot 3)\cdot (x^{\,2}\cdot x\,)\cdot y^{\,2}\cdot z^{\, 4}+\\\kern{14em} +5\cdot (x^{\,2}\cdot x\,)\cdot (\,y^{\,2}\cdot y\,)\cdot z-(5\cdot 11)\cdot (x^{\,2}\cdot x\,)\cdot y^{\,2}\cdot z^{\,3}=\\\kern{5em} =30\cdot x^{\,2+3}\cdot y^{\,2+5}\cdot z+15\cdot x^{\,2+1}\cdot y^{\,2}\cdot z^{\, 4}+5\cdot x^{\,2+1}\cdot y^{\,2+1}\cdot z-55\cdot x^{\,2+1}\cdot y^{\,2}\cdot z^{\,3}=\\\kern{18em} =30x^{\,5}y^{\,7}z+15x^{\,3}y^{\,2}z^{\, 4}+5x^{\,3}y^{\,3}z-55x^{\,3}y^{\,2}z^{\,3}{\small .}\end{array}\)

Осылайша,

\(\displaystyle 5x^{\,2}y^{\,2}\cdot (6x^{\,3}y^{\,5}z+3xz^{\, 4}+xyz-11xz^{\,3})=30x^{\,5}y^{\,7}z+15x^{\,3}y^{\,2}z^{\, 4}+5x^{\,3}y^{\,3}z-55x^{\,3}y^{\,2}z^{\,3}{\small .}\)

Жауабы: \(\displaystyle 30x^{\,5}y^{\,7}z+15x^{\,3}y^{\,2}z^{\, 4}+5x^{\,3}y^{\,3}z-55x^{\,3}y^{\,2}z^{\,3}{\small .}\)