290 Вправа 290Алгебра 8 клас Бевз 2025 · § 6 · стор. 60

290. Поділи многочлен на многочлен, записавши частку у вигляді дробу і скоротивши його.

а) ${\displaystyle (a^5-a):(a^2-1)=}$${\displaystyle \frac{a^5-a}{a^2-1}}$=

=${\displaystyle \frac{a\cancel{\left(a+1\right)}(a^2+1)\cancel{\left(a-1\right)}}{\cancel{\left(a+1\right)}\cancel{\left(a-1\right)}}=}$${\displaystyle a\left(a^2+1\right)}$;

б) ${\displaystyle (z^6-z^2):(z^2-1)=}$${\displaystyle \frac{z^6-z^2}{z^2-1}}$=

=${\displaystyle \frac{z^2\cancel{\left(z+1\right)}(z^2+1)\cancel{\left(z-1\right)}}{\cancel{\left(z+1\right)}\cancel{\left(z-1\right)}}=}$${\displaystyle z^2\left(z^2+1\right)}$;

в) ${\displaystyle (2a+8b):(16b^2-a^2)=}$${\displaystyle \frac{2a+8b}{16b^2-a^2}}$=

=${\displaystyle -\frac{2\cancel{\left(a+4b\right)}}{\cancel{\left(a+4b\right)}(a-4b)}=}$${\displaystyle -\frac{2}{a-4b}=}$${\displaystyle \frac{2}{4b-a}}$;

г) ${\displaystyle (4x^2-4xy+y^2):(y^2-4x^2)=}$${\displaystyle \frac{4x^2-4xy+y^2}{y^2-4x^2}}$=

=${\displaystyle -\frac{\cancel{\left(2x-y\right)}(2x-y)}{(2x+y)\cancel{\left(2x-y\right)}}=}$${\displaystyle -\frac{2x-y}{2x+y}=}$${\displaystyle \frac{y-2x}{y+2x}}$.