Теориясы: Көпмүшені көпмүшеге көбейту

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

\(\displaystyle \begin{array}{l}(\color{blue}{2x^{\,3}y^{\,3}z^{\,4}}-\color{green}{3x^{\,4}y^{\,5}z}+\color{red}{4x^{\,2}yz^{\,7}})\cdot (2x^{\,3}y^{\,3}+3x^{\,2}yz^{\,3})=\\\kern{8em} =\color{blue}{2x^{\,3}y^{\,3}z^{\,4}}\cdot (2x^{\,3}y^{\,3}+3x^{\,2}yz^{\,3})-\color{green}{3x^{\,4}y^{\,5}z} \cdot (2x^{\,3}y^{\,3}+3x^{\,2}yz^{\,3})+\\\kern{24em} +\color{red}{4x^{\,2}yz^{\,7}}\cdot (2x^{\,3}y^{\,3}+3x^{\,2}yz^{\,3}){\small .}\end{array}\)

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

\(\displaystyle \begin{array}{l}\color{blue}{2x^{\,3}y^{\,3}z^{\,4}}\cdot (2x^{\,3}y^{\,3}+3x^{\,2}yz^{\,3})-\color{green}{3x^{\,4}y^{\,5}z}\cdot (2x^{\,3}y^{\,3}+3x^{\,2}yz^{\,3})+\color{red}{4x^{\,2}yz^{\,7}}\cdot (2x^{\,3}y^{\,3}+3x^{\,2}yz^{\,3})=\\\kern{2em} =\color{blue}{2x^{\,3}y^{\,3}z^{\,4}}\cdot 2x^{\,3}y^{\,3}+\color{blue}{2x^{\,3}y^{\,3}z^{\,4}}\cdot 3x^{\,2}yz^{\,3}-(\color{green}{3x^{\,4}y^{\,5}z}\cdot 2x^{\,3}y^{\,3}+\color{green}{3x^{\,4}y^{\,5}z}\cdot 3x^{\,2}yz^{\,3})+\\\kern{21em} +(\color{red}{4x^{\,2}yz^{\,7}}\cdot 2x^{\,3}y^{\,3}+\color{red}{4x^{\,2}yz^{\,7}}\cdot 3x^{\,2}yz^{\,3})=\\\kern{2em} =(2\cdot 2)\cdot (x^{\,3}\cdot x^{\,3})\cdot (\,y^{\,3}\cdot y^{\,3})\cdot z^{\,4}+(2\cdot 3)\cdot (x^{\,3}\cdot x^{\,2})\cdot (\,y^{\,3}\cdot y\,)\cdot (z^{\,4}\cdot z^{\,3})-\\\kern{8em} -\big((3\cdot 2)\cdot (x^{\,4}\cdot x^{\,3})\cdot (\,y^{\,5}\cdot y^{\,3})\cdot z+(3\cdot 3)\cdot (x^{\,4}\cdot x^{\,2})\cdot (\,y^{\,5}\cdot y\,)\cdot (z\cdot z^{\,3})\big)+\\\kern{8em} +\big((4\cdot 2)\cdot (x^{\,2}\cdot x^{\,3})\cdot (\,y\cdot y^{\,3})\cdot z^{\,7}+(4\cdot 3)\cdot (x^{\,2}\cdot x^{\,2})\cdot (\,y\cdot y\,)\cdot (z^{\,7}\cdot z^{\,3})\big)=\\\kern{2em} =4\cdot x^{\,3+3}\cdot y^{\,3+3}\cdot z^{\,4}+6\cdot x^{\,3+2}\cdot y^{\,3+1}\cdot z^{\,4+3}-(6\cdot x^{\,4+3}\cdot y^{\,5+3}\cdot z+9\cdot x^{\,4+2}\cdot y^{\,5+1}\cdot z^{\,1+3})+\\\kern{18em} +(8\cdot x^{\,2+3}\cdot y^{\,1+3}\cdot z^{\,7}+12\cdot x^{\,2+2}\cdot y^{\,1+1}\cdot z^{\,7+3})=\\\kern{7em} =4x^{\,6}y^{\,6}z^{\,4}+6x^{\,5}y^{\,4}z^{\,7}-(6x^{\,7}y^{\,8}z+9x^{\,6}y^{\,6}z^{\,4})+(8x^{\,5}y^{\,4}z^{\,7}+12x^{\,4}y^{\,2}z^{\,10}){\small .}\end{array}\)

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

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

Алынған көпмүшені ұқсас бірмүшелерді келтіре отырып, стандарт түрге келтірейік:

\(\displaystyle \begin{array}{l}4\color{blue}{x^{\,6}y^{\,6}z^{\,4}}+6\color{green}{x^{\,5}y^{\,4}z^{\,7}}-6x^{\,7}y^{\,8}z-9\color{blue}{x^{\,6}y^{\,6}z^{\,4}}+8\color{green}{x^{\,5}y^{\,4}z^{\,7}}+12x^{\,4}y^{\,2}z^{\,10}=\\\kern{4em} =(4\color{blue}{x^{\,6}y^{\,6}z^{\,4}}-9\color{blue}{x^{\,6}y^{\,6}z^{\,4}})+(6\color{green}{x^{\,5}y^{\,4}z^{\,7}}+8\color{green}{x^{\,5}y^{\,4}z^{\,7}})-6x^{\,7}y^{\,8}z+12x^{\,4}y^{\,2}z^{\,10}=\\\kern{4em} =(4-9)\color{blue}{x^{\,6}y^{\,6}z^{\,4}}+(6+8)\color{green}{x^{\,5}y^{\,4}z^{\,7}}-6x^{\,7}y^{\,8}z+12x^{\,4}y^{\,2}z^{\,10}=\\\kern{16em} =-5\color{blue}{x^{\,6}y^{\,6}z^{\,4}}+14\color{green}{x^{\,5}y^{\,4}z^{\,7}}-6x^{\,7}y^{\,8}z+12x^{\,4}y^{\,2}z^{\,10}{\small .}\end{array}\)

Осылайша,

\(\displaystyle \begin{array}{l}(2x^{\,3}y^{\,3}z^{\,4}-3x^{\,4}y^{\,5}z+4x^{\,2}yz^{\,7})(2x^{\,3}y^{\,3}+3x^{\,2}yz^{\,3})=\\\kern{15em} =-5x^{\,6}y^{\,6}z^{\,4}+14x^{\,5}y^{\,4}z^{\,7}-6x^{\,7}y^{\,8}z+12x^{\,4}y^{\,2}z^{\,10}{\small .}\end{array}\)

Жауабы: \(\displaystyle -5x^{\,6}y^{\,6}z^{\,4}+14x^{\,5}y^{\,4}z^{\,7}-6x^{\,7}y^{\,8}z+12x^{\,4}y^{\,2}z^{\,10}{\small .}\)