Теориясы: Көпмүшені бірмүшеге көбейту

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

• \(\displaystyle -y^{\,2}z\cdot 8xz\cdot 3yz^{\,3}=-(8\cdot 3)\cdot x\cdot (\,y^{\,2}\cdot y\,) \cdot (z\cdot z\cdot z^{\,3})=-24\cdot x\cdot y^{\,2+1}\cdot z^{\,1+1+3}=-24xy^{\,3}z^{\,5}{\small ,}\)
• \(\displaystyle 3yz\cdot 2xyz^{\,2}=(3\cdot 2)\cdot x\cdot (\,y\cdot y\,)\cdot (z\cdot z^{\,2})=6\cdot x\cdot y^{\,1+1}\cdot z^{\,1+2}=6xy^{\,2}z^{\,3}{\small .}\)

Сонда

\(\displaystyle \begin{array}{l}-5xy^{\,2}z^{\,3}(-y^{\,2}z\cdot 8xz\cdot 3yz^{\,3}+5xyz^{\,3}+4yz^{\,2}(3yz\cdot 2xyz^{\,2}-xz+xyz\,))=\\\kern{10em} =-5xy^{\,2}z^{\,3}(-24xy^{\,3}z^{\,5}+5xyz^{\,3}+4yz^{\,2}(6xy^{\,2}z^{\,3}-xz+xyz\,)){\small .}\end{array}\)

2. . Әрбір қосылғышты \(\displaystyle 4yz^{\,2}\) көбейте отырып және алынған әрбір бірмүшені стандарт түрге түрлендіре отырып, ішкі жақшаларды ашайық:

\(\displaystyle \begin{array}{l}\color{blue}{4yz^{\,2}}(6xy^{\,2}z^{\,3}-xz+xyz\,)=\color{blue}{4yz^{\,2}}\cdot 6xy^{\,2}z^{\,3}-\color{blue}{4yz^{\,2}}\cdot xz+\color{blue}{4yz^{\,2}}\cdot xyz=\\\;\phantom{\color{blue}{4yz^{\,2}}(6xy^{\,2}z^{\,3}-xz+xyz\,)}=(4\cdot 6)\cdot x\cdot (\,y\cdot y^{\,2})\cdot (z^{\,2}\cdot z^{\,3})-4\cdot x\cdot y\cdot (z^{\,2}\cdot z\,)+\\\kern{23em} +4\cdot x\cdot (\,y\cdot y\,)\cdot (z^{\,2}\cdot z\,)=\\\;\phantom{\color{blue}{4yz^{\,2}}(6xy^{\,2}z^{\,3}-xz+xyz\,)}=24\cdot x\cdot y^{\,1+2}\cdot z^{\,2+3}-4\cdot x\cdot y\cdot z^{\,2+1}+4\cdot x\cdot y^{\,1+1}\cdot z^{\,2+1}=\\\;\phantom{\color{blue}{4yz^{\,2}}(6xy^{\,2}z^{\,3}-xz+xyz\,)}=24xy^{\,3}z^{\,5}-4xyz^{\,3}+4xy^{\,2}z^{\,3}{\small .}\end{array}\)

Сондықтан

\(\displaystyle \begin{array}{l}-5xy^{\,2}z^{\,3}(-24xy^{\,3}z^{\,5}+5xyz^{\,3}+4yz^{\,2}(6xy^{\,2}z^{\,3}-xz+xyz\,))=\\\kern{11em} =-5xy^{\,2}z^{\,3}(-24xy^{\,3}z^{\,5}+5xyz^{\,3}+24xy^{\,3}z^{\,5}-4xyz^{\,3}+4xy^{\,2}z^{\,3}){\small .}\end{array}\)

3. Жақшадағы көпмүшені стандарт түрге түрлендіріп, ұқсас бірмүшелерді келтіреміз:

\(\displaystyle \begin{array}{l}-5xy^{\,2}z^{\,3}(-24\color{blue}{xy^{\,3}z^{\,5}}+5\color{green}{xyz^{\,3}}+24\color{blue}{xy^{\,3}z^{\,5}}-4\color{green}{xyz^{\,3}}+4xy^{\,2}z^{\,3})=\\\kern{5em} =-5xy^{\,2}z^{\,3}((-24\color{blue}{xy^{\,3}z^{\,5}}+24\color{blue}{xy^{\,3}z^{\,5}})+(5\color{green}{xyz^{\,3}}-4\color{green}{xyz^{\,3}})+4xy^{\,2}z^{\,3})=\\\kern{5em} =-5xy^{\,2}z^{\,3}((-24+24)\color{blue}{xy^{\,3}z^{\,5}}+(5-4)\color{green}{xyz^{\,3}}+4xy^{\,2}z^{\,3})=\\\kern{5em} =-5xy^{\,2}z^{\,3}(0\cdot \color{blue}{xy^{\,3}z^{\,5}}+\color{green}{xyz^{\,3}}+4xy^{\,2}z^{\,3})=-5xy^{\,2}z^{\,3}(\color{green}{xyz^{\,3}}+4xy^{\,2}z^{\,3}){\small .}\end{array}\)

4. Тағы да жақшадағы әрбір қосылғышты \(\displaystyle -5xy^{\,2}z^{\,3}{\small }\) көбейту арқылы жақшаларды ашайық:

\(\displaystyle \begin{array}{l}\color{blue}{-5xy^{\,2}z^{\,3}}\cdot (xyz^{\,3}+4xy^{\,2}z^{\,3})=\\\kern{7em} =(\color{blue}{-5xy^{\,2}z^{\,3}})\cdot xyz^{\,3}+(\color{blue}{-5xy^{\,2}z^{\,3}})\cdot 4xy^{\,2}z^{\,3}=\\\kern{7em} =-5\cdot (x\cdot x\,)\cdot (\,y^{\,2}\cdot y\,)\cdot (z^{\,3}\cdot z^{\,3})+((-5)\cdot 4)\cdot (x\cdot x\,)\cdot (\,y^{\,2}\cdot y^{\,2})\cdot (z^{\,3}\cdot z^{\,3})=\\\kern{7em} =-5\cdot x^{\,1+1}\cdot y^{\,2+1}\cdot z^{\,3+3}-20\cdot x^{\,1+1}\cdot y^{\,2+2}\cdot z^{\,3+3}=\\\kern{21em} =-5x^{\,2}y^{\,3}z^{\,6}-20x^{\,2}y^{\,4}z^{\,6}{\small .}\end{array}\)

Осылайша,

\(\displaystyle \begin{array}{l}-5xy^{\,2}z^{\,3}(-y^{\,2}z\cdot 8xz\cdot 3yz^{\,3}+5xyz^{\,3}+4yz^{\,2}(3yz\cdot 2xyz^{\,2}-xz+xyz\,))=\\\kern{24em} =-5x^{\,2}y^{\,3}z^{\,6}-20x^{\,2}y^{\,4}z^{\,6}{\small .}\end{array}\)

Жауабы: \(\displaystyle -5x^{\,2}y^{\,3}z^{\,6}-20x^{\,2}y^{\,4}z^{\,6}{\small .}\)