291 Вправа 291Алгебра 8 клас Бевз 2025 · § 6 · стор. 60

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

а) ${\displaystyle (c^6-c^2):(c^3-c)=}$${\displaystyle \frac{c^6-c^2}{c^3-c}}$=

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

б) ${\displaystyle (c^4+c):(c^2+c)=}$${\displaystyle \frac{c^4+c}{c^2+c}}$=

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

в) ${\displaystyle (x^3+4x^2+4x):(x^2+2x)=}$${\displaystyle \frac{x^3+4x^2+4x}{x^2+2x}}$=

=${\displaystyle \frac{\cancel{x}\cancel{\left(x+2\right)}(x+2)}{\cancel{x}\cancel{\left(x+2\right)}}=}$${\displaystyle x+2}$;

г) ${\displaystyle (x^4+x):(x^2-x+1)=}$${\displaystyle \frac{x^4+x}{x^2-x+1}}$=

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