Көпмүшені бірмүшеге бөлу кезінде бөліндіні табыңыз:
Жауаптағы сандық бөлшек коэффициенттерді жай бөлшектер түрінде жазыңыз.
\(\displaystyle 3x^{\,15}y^{\,7}z^{\,5}-\frac{3}{8}x^{\,11}y^{\,10}z^{\,6}+\frac{1}{3}x^{\,7}y^{\,6}z^{\,6}+\frac{5}{6}x^{\,4}y^{\,4}z^{\,3}\) көпмүшесін \(\displaystyle \frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\) бөлу оның әр мүшесін \(\displaystyle \frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\) бөлуді білдіреді. Сондықтан:
\(\displaystyle \begin{array}{l}\left(3x^{\,15}y^{\,7}z^{\,5}-\frac{3}{8}x^{\,11}y^{\,10}z^{\,6}+\frac{1}{3}x^{\,7}y^{\,6}z^{\,6}+\frac{5}{6}x^{\,4}y^{\,4}z^{\,3}\right):\left(\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\right)=\\\kern{3em} =\left(3x^{\,15}y^{\,7}z^{\,5}\right):\left(\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\right)-\left(\frac{3}{8}x^{\,11}y^{\,10}z^{\,6}\right):\left(\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\right)+\\\kern{11em} +\left(\frac{1}{3}x^{\,7}y^{\,6}z^{\,6}\right):\left(\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\right)+ \left(\frac{5}{6}x^{\,4}y^{\,4}z^{\,3}\right):\left(\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\right){\small .}\end{array}\)
\(\displaystyle ":"\) бөлу белгісін бөлшек сызығына алмастырайық:
\(\displaystyle \begin{array}{l}\left(3x^{\,15}y^{\,7}z^{\,5}\right):\left(\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\right)-\left(\frac{3}{8}x^{\,11}y^{\,10}z^{\,6}\right):\left(\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\right)+\\\kern{11em} +\left(\frac{1}{3}x^{\,7}y^{\,6}z^{\,6}\right):\left(\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\right)+ \left(\frac{5}{6}x^{\,4}y^{\,4}z^{\,3}\right):\left(\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}\right)=\\[10px]\kern{15em} =\frac{3x^{\,15}y^{\,7}z^{\,5}}{\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}}-\frac{\frac{3}{8}x^{\,11}y^{\,10}z^{\,6}}{\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}}+\frac{\frac{1}{3}x^{\,7}y^{\,6}z^{\,6}}{\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}}+\frac{\frac{5}{6}x^{\,4}y^{\,4}z^{\,3}}{\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}}{\small .}\end{array}\)
Әр бөлшектегі сандық коэффициенттерді бір-біріне бөлейік және дәрежелерге дәрежелер бөліндісі формуласын қолданайық:
\(\displaystyle \begin{array}{l}\frac{3x^{\,15}y^{\,7}z^{\,5}}{\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}}-\frac{\frac{3}{8}x^{\,11}y^{\,10}z^{\,6}}{\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}}+\frac{\frac{1}{3}x^{\,7}y^{\,6}z^{\,6}}{\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}}+\frac{\frac{5}{6}x^{\,4}y^{\,4}z^{\,3}}{\frac{3}{7}x^{\,2}y^{\,4}z^{\,2}}=\\[10px]\kern{5em} =\left(3:\frac{3}{7}\right)x^{\,15-2}y^{\,7-4}z^{\,5-2}-\left(\frac{3}{8}:\frac{3}{7}\right)x^{\,11-2}y^{\,10-4}z^{\,6-2}+\\\kern{11em} +\left(\frac{1}{3}:\frac{3}{7}\right)x^{\,7-2}y^{\,6-4}z^{\,6-2}+\left(\frac{5}{6}:\frac{3}{7}\right)x^{\,4-2}y^{\,4-4}z^{\,3-2}=\\[10px]\kern{16em} =7x^{\,13}y^{\,3}z^{\,3}-\frac{7}{8}x^{\,9}y^{\,6}z^{\,4}+\frac{7}{9}x^{\,5}y^{\,2}z^{\,4}+\frac{35}{18}x^{\,2}z{\small .}\end{array}\)
Жауабы: \(\displaystyle 7x^{\,13}y^{\,3}z^{\,3}-\frac{7}{8}x^{\,9}y^{\,6}z^{\,4}+\frac{7}{9}x^{\,5}y^{\,2}z^{\,4}+\frac{35}{18}x^{\,2}z{\small .}\)