282 Вправа 282Алгебра 8 клас Бевз 2025 · § 6 · стор. 59

282. Виконай ділення.

а) ${\displaystyle \frac{x^2-6x+9}{4x^2-6x+9}:\frac{6-2x}{27+8x^3}=}$${\displaystyle \frac{x^2-6x+9}{4x^2-6x+9}\cdot \frac{27+8x^3}{6-2x}}$=

=${\displaystyle \frac{(x^2-6x+9)\cdot (27+8x^3)}{(4x^2-6x+9)\cdot (6-2x)}}$=

=${\displaystyle -\frac{(2x+3)\cancel{\left(4x^2-6x+9\right)}\cancel{\left(x-3\right)}(x-3)}{2\cancel{\left(4x^2-6x+9\right)}\cancel{\left(x-3\right)}}}$=

=${\displaystyle -\frac{(x-3)(2x+3)}{2}=}$${\displaystyle \frac{\left(3-x\right)\left(3+2x\right)}{2}}$;

б) ${\displaystyle \frac{a^2+ax+x^2}{x-1}:\frac{a^3-x^3}{x^2-1}=}$${\displaystyle \frac{a^2+ax+x^2}{x-1}\cdot \frac{x^2-1}{a^3-x^3}}$=

=${\displaystyle \frac{(a^2+ax+x^2)\cdot (x^2-1)}{(x-1)\cdot (a^3-x^3)}}$=

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