Теориясы: Көпмүшені көпмүшеге көбейту

Жақшаларды көбейту үшін алдымен бірінші жақшаның әр мүшесін екінші жақшаға көбейтеміз:

\(\displaystyle \begin{array}{l}(\color{blue}{x^{\,5}y^{\,2}}-\color{green}{2x^{\,7}yz^{\,4}}+\color{red}{3y^{\,6}z^{\,3}})\cdot (2x^{\,2}y^{\,3}z^{\,8}+3y^{\,4}z^{\,4}+3x^{\,5}z\,)=\\\kern{6em} =\color{blue}{x^{\,5}y^{\,2}}\cdot (2x^{\,2}y^{\,3}z^{\,8}+3y^{\,4}z^{\,4}+3x^{\,5}z\,)-\color{green}{2x^{\,7}yz^{\,4}} \cdot (2x^{\,2}y^{\,3}z^{\,8}+3y^{\,4}z^{\,4}+3x^{\,5}z\,)+\\\kern{22em} +\color{red}{3y^{\,6}z^{\,3}}\cdot (2x^{\,2}y^{\,3}z^{\,8}+3y^{\,4}z^{\,4}+3x^{\,5}z\,){\small .}\end{array}\)

Әрі қарай әр жақшаны олардың алдындағы көбейткішке көбейтіп, алынған бірмүшелерді стандарт түрге келтіреміз:

\(\displaystyle \begin{array}{l}\color{blue}{x^{\,5}y^{\,2}}\cdot (2x^{\,2}y^{\,3}z^{\,8}+3y^{\,4}z^{\,4}+3x^{\,5}z\,)-\color{green}{2x^{\,7}yz^{\,4}}\cdot (2x^{\,2}y^{\,3}z^{\,8}+3y^{\,4}z^{\,4}+3x^{\,5}z\,)+\\\kern{22em} +\color{red}{3y^{\,6}z^{\,3}}\cdot (2x^{\,2}y^{\,3}z^{\,8}+3y^{\,4}z^{\,4}+3x^{\,5}z\,)=\\\kern{3em} =\color{blue}{x^{\,5}y^{\,2}}\cdot 2x^{\,2}y^{\,3}z^{\,8}+\color{blue}{x^{\,5}y^{\,2}}\cdot 3y^{\,4}z^{\,4}+\color{blue}{x^{\,5}y^{\,2}}\cdot 3x^{\,5}z-\\\kern{14em} -(\color{green}{2x^{\,7}yz^{\,4}}\cdot 2x^{\,2}y^{\,3}z^{\,8}+\color{green}{2x^{\,7}yz^{\,4}}\cdot 3y^{\,4}z^{\,4}+\color{green}{2x^{\,7}yz^{\,4}}\cdot 3x^{\,5}z\,)+\\\kern{14em} +(\color{red}{3y^{\,6}z^{\,3}}\cdot 2x^{\,2}y^{\,3}z^{\,8}+\color{red}{3y^{\,6}z^{\,3}}\cdot 3y^{\,4}z^{\,4}+\color{red}{3y^{\,6}z^{\,3}}\cdot 3x^{\,5}z\,)=\\\kern{3em} =2\cdot (x^{\,5}\cdot x^{\,2})\cdot (\,y^{\,2}\cdot y^{\,3})\cdot z^{\,8}+3\cdot x^{\,5}\cdot (\,y^{\,2}\cdot y^{\,4})\cdot z^{\,4}+3\cdot (x^{\,5}\cdot x^{\,5})\cdot y^{\,2}\cdot z\,-\\\kern{7em} -\big((2\cdot 2)\cdot (x^{\,7}\cdot x^{\,2})\cdot (\,y\cdot y^{\,3})\cdot (z^{\,4}\cdot z^{\,8})+(2\cdot 3)\cdot x^{\,7}\cdot (\,y\cdot y^{\,4})\cdot (z^{\,4}\cdot z^{\,4})+\\\kern{7em} +(2\cdot 3)\cdot (x^{\,7}\cdot x^{\,5})\cdot y\cdot (z^{\,4}\cdot z\,)\big)+\big((3\cdot 2)\cdot x^{\,2}\cdot (\,y^{\,6}\cdot y^{\,3})\cdot (z^{\,3}\cdot z^{\,8})+\\\kern{7em} +(3\cdot 3)\cdot (\,y^{\,6}\cdot y^{\,4})\cdot (z^{\,3}\cdot z^{\,4})+(3\cdot 3)\cdot x^{\,5}\cdot y^{\,6}\cdot (z^{\,3}\cdot z\,)\big)=\\\kern{3em} =2\cdot x^{\,5+2}\cdot y^{\,2+3}\cdot z^{\,8}+3\cdot x^{\,5}\cdot y^{\,2+4}\cdot z^{\,4}+3\cdot x^{\,5+5}\cdot y^{\,2}\cdot z\,-\\\kern{10em} -(4\cdot x^{\,7+2}\cdot y^{\,1+3}\cdot z^{\,4+8}+6\cdot x^{\,7}\cdot y^{\,1+4}\cdot z^{\,4+4}+6\cdot x^{\,7+5}\cdot y\cdot z^{\,4+1})+\\\kern{10em} +(6\cdot x^{\,2}\cdot y^{\,6+3}\cdot z^{\,3+8}+9\cdot y^{\,6+4}\cdot z^{\,3+4}+9\cdot x^{\,5}\cdot y^{\,6}\cdot z^{\,3+1})=\\\kern{8em} =2x^{\,7}y^{\,5}z^{\,8}+3x^{\,5}y^{\,6}z^{\,4}+3x^{\,10}y^{\,2}z-(4x^{\,9}y^{\,4}z^{\,12}+6x^{\,7}y^{\,5}z^{\,8}+6x^{\,12}yz^{\,5})+\\\kern{23em} +(6x^{\,2}y^{\,9}z^{\,11}+9y^{\,10}z^{\,7}+9x^{\,5}y^{\,6}z^{\,4}){\small .}\end{array}\)

Жақшаларды ашайық. Жақшалардың алдында минус таңбасы тұрғандықтан, осы жақшалардың ішіндегі барлық таңбалар қарама-қарсы таңбаға өзгереді:

\(\displaystyle \begin{array}{l}2x^{\,7}y^{\,5}z^{\,8}+3x^{\,5}y^{\,6}z^{\,4}+3x^{\,10}y^{\,2}z-(4x^{\,9}y^{\,4}z^{\,12}+6x^{\,7}y^{\,5}z^{\,8}+6x^{\,12}yz^{\,5})+\\\kern{23em} +(6x^{\,2}y^{\,9}z^{\,11}+9y^{\,10}z^{\,7}+9x^{\,5}y^{\,6}z^{\,4})=\\\kern{5em} =2x^{\,7}y^{\,5}z^{\,8}+3x^{\,5}y^{\,6}z^{\,4}+3x^{\,10}y^{\,2}z-4x^{\,9}y^{\,4}z^{\,12}-6x^{\,7}y^{\,5}z^{\,8}-6x^{\,12}yz^{\,5}+\\\kern{23em} +6x^{\,2}y^{\,9}z^{\,11}+9y^{\,10}z^{\,7}+9x^{\,5}y^{\,6}z^{\,4}{\small .}\end{array}\)

Алынған көпмүшені ұқсас бірмүшелерді келтіре отырып, стандарт түрге келтірейік:

\(\displaystyle \begin{array}{l}2\color{blue}{x^{\,7}y^{\,5}z^{\,8}}+3\color{green}{x^{\,5}y^{\,6}z^{\,4}}+3x^{\,10}y^{\,2}z-4x^{\,9}y^{\,4}z^{\,12}-6\color{blue}{x^{\,7}y^{\,5}z^{\,8}}-6x^{\,12}yz^{\,5}+\\\kern{23em} +6x^{\,2}y^{\,9}z^{\,11}+9y^{\,10}z^{\,7}+9\color{green}{x^{\,5}y^{\,6}z^{\,4}}=\\\kern{3em} =(2\color{blue}{x^{\,7}y^{\,5}z^{\,8}}-6\color{blue}{x^{\,7}y^{\,5}z^{\,8}})+(3\color{green}{x^{\,5}y^{\,6}z^{\,4}}+9\color{green}{x^{\,5}y^{\,6}z^{\,4}})+3x^{\,10}y^{\,2}z\,-\\\kern{18em} -4x^{\,9}y^{\,4}z^{\,12}-6x^{\,12}yz^{\,5}+6x^{\,2}y^{\,9}z^{\,11}+9y^{\,10}z^{\,7}=\\\kern{3em} =(2-6)\color{blue}{x^{\,7}y^{\,5}z^{\,8}}+(3+9)\color{green}{x^{\,5}y^{\,6}z^{\,4}}+3x^{\,10}y^{\,2}z-4x^{\,9}y^{\,4}z^{\,12}-\\\kern{23em} -6x^{\,12}yz^{\,5}+6x^{\,2}y^{\,9}z^{\,11}+9y^{\,10}z^{\,7}=\\\kern{3em} =-4\color{blue}{x^{\,7}y^{\,5}z^{\,8}}+12\color{green}{x^{\,5}y^{\,6}z^{\,4}}+3x^{\,10}y^{\,2}z-4x^{\,9}y^{\,4}z^{\,12}-6x^{\,12}yz^{\,5}+6x^{\,2}y^{\,9}z^{\,11}+9y^{\,10}z^{\,7}{\small .}\end{array}\)

Осылайша,

\(\displaystyle \begin{array}{l}(x^{\,5}y^{\,2}-2x^{\,7}yz^{\,4}+3y^{\,6}z^{\,3})(2x^{\,2}y^{\,3}z^{\,8}+3y^{\,4}z^{\,4}+3x^{\,5}z\,)=\\\kern{3em} =-4x^{\,7}y^{\,5}z^{\,8}+12x^{\,5}y^{\,6}z^{\,4}+3x^{\,10}y^{\,2}z-4x^{\,9}y^{\,4}z^{\,12}-6x^{\,12}yz^{\,5}+6x^{\,2}y^{\,9}z^{\,11}+9y^{\,10}z^{\,7}{\small .}\end{array}\)

Жауабы: \(\displaystyle -4x^{\,7}y^{\,5}z^{\,8}+12x^{\,5}y^{\,6}z^{\,4}+3x^{\,10}y^{\,2}z-4x^{\,9}y^{\,4}z^{\,12}-6x^{\,12}yz^{\,5}+6x^{\,2}y^{\,9}z^{\,11}+9y^{\,10}z^{\,7}{\small .}\)