Көпмүшелердің айырмасын табыңыз:
Жауапта көпмүшені стандарт түрде жазыңыз.
Бізге берілген көпмүшелерді стандарт түрге келтірейік:
\(\displaystyle \quad \begin{array}{l}{\small 1)\,}3u^{\,3}s^{\,2}t\cdot 7u^{\,2}\cdot st+3s^{\,4}t^{\,3}\cdot u^{\,5}s^{\,2}t-10ust=\\\kern{5em} =(3\cdot 7)\cdot (u^{\,3}\cdot u^{\,2})\cdot (s^{\,2}\cdot s\,)\cdot (t\cdot t\,)+3\cdot u^{\,5}\cdot (s^{\,4}\cdot s^{\,2})\cdot (t^{\,3}\cdot t\,)-10ust=\\\kern{5em} =21\cdot u^{\,3+2}\cdot s^{\,2+1}\cdot t^{\,1+1}+3\cdot u^{\,5}\cdot s^{\,4+2}\cdot t^{\,3+1}-10ust=\\\kern{20em} =21u^{\,5}s^{\,3}t^{\,2}+3u^{\,5}s^{\,6}t^{\,4}-10ust\,{\small ;}\end{array}\)
\(\displaystyle \quad \begin{array}{l}{\small 2)\,}4u^{\,2}t\cdot (-3)us^{\,3}-10ust+s^{\,3}t^{\,3}\cdot u^{\,2}\cdot s^{\,3}t\cdot 3u^{\,3}=\\\kern{5em} =(4\cdot (-3))\cdot (u^{\,2}\cdot u\,)\cdot s^{\,3}\cdot t-10ust+3\cdot (u^{\,2}\cdot u^{\,3})\cdot (s^{\,3}\cdot s^{\,3})\cdot (t^{\,3}\cdot t\,)=\\\kern{5em} =-12\cdot u^{\,2+1}\cdot s^{\,3}\cdot t-10ust+3\cdot u^{\,2+3}\cdot s^{\,3+3}\cdot t^{\,3+1}=\\\kern{19em} =-12u^{\,3}s^{\,3}t-10ust+3u^{\,5}s^{\,6}t^{\,4}{\small .}\end{array}\)
Демек, бізге көпмүшелерді азайту қажет
\(\displaystyle 21u^{\,5}s^{\,3}t^{\,2}+3u^{\,5}s^{\,6}t^{\,4}-10ust\) және \(\displaystyle -12u^{\,3}s^{\,3}t-10ust+3u^{\,5}s^{\,6}t^{\,4}{\small .}\)
Олардың айырмасын жазайық:
\(\displaystyle (\color{blue}{21u^{\,5}s^{\,3}t^{\,2}+3u^{\,5}s^{\,6}t^{\,4}-10ust}\,)-(\color{green}{-12u^{\,3}s^{\,3}t-10ust+3u^{\,5}s^{\,6}t^{\,4}}\,){\small .}\)
Жақшаларды ашайық. Жақшаның алдында минус таңбасы тұрғандықтан, жақшаның ішіндегі барлық таңбалар қарама-қарсы болып өзгереді:
\(\displaystyle \begin{array}{l}(21u^{\,5}s^{\,3}t^{\,2}+3u^{\,5}s^{\,6}t^{\,4}-10ust\,)-(-12u^{\,3}s^{\,3}t-10ust+3u^{\,5}s^{\,6}t^{\,4}\,)=\\\kern{5em} =21u^{\,5}s^{\,3}t^{\,2}+3u^{\,5}s^{\,6}t^{\,4}-10ust+12u^{\,3}s^{\,3}t+10ust-3u^{\,5}s^{\,6}t^{\,4}{\small .}\end{array}\)
Ұқсас мүшелерді келтірейік:
\(\displaystyle \begin{array}{l}21u^{\,5}s^{\,3}t^{\,2}+3\color{blue}{u^{\,5}s^{\,6}t^{\,4}}-10\color{green}{ust}+12u^{\,3}s^{\,3}t+10\color{green}{ust}-3\color{blue}{u^{\,5}s^{\,6}t^{\,4}}=\\\kern{2em} =21u^{\,5}s^{\,3}t^{\,2}+(3\color{blue}{u^{\,5}s^{\,6}t^{\,4}}-3\color{blue}{u^{\,5}s^{\,6}t^{\,4}})+(-10\color{green}{ust}+10\color{green}{ust}\,)+12u^{\,3}s^{\,3}t=\\\kern{2em} =21u^{\,5}s^{\,3}t^{\,2}+(3-3)\color{blue}{u^{\,5}s^{\,6}t^{\,4}}+(-10+10)\color{green}{ust}+12u^{\,3}s^{\,3}t=\\\kern{6em} =21u^{\,5}s^{\,3}t^{\,2}+0\cdot\color{blue}{u^{\,5}s^{\,6}t^{\,4}}+0\cdot\color{green}{ust}+12u^{\,3}s^{\,3}t=21u^{\,5}s^{\,3}t^{\,2}+12u^{\,3}s^{\,3}t{\small .}\end{array}\)
Жауабы: \(\displaystyle 21u^{\,5}s^{\,3}t^{\,2}+12u^{\,3}s^{\,3}t{\small .}\)