310 Вправа 310Алгебра 8 клас Бевз 2025 · § 7 · стор. 65

310. Спрости вираз.

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

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

б) ${\displaystyle {\left(\frac{2a}{a-x}\right)}^3\cdot \frac{a^2-2ax+x^2}{6a^2}}$;

1) ${\displaystyle {\left(\frac{2a}{a-x}\right)}^3=}$${\displaystyle \frac{{\left(2a\right)}^3}{{(a-x)}^3}=}$${\displaystyle \frac{8a^3}{{(a-x)}^3}}$;

2) ${\displaystyle \frac{8a^3}{{(a-x)}^3}\cdot \frac{a^2-2ax+x^2}{6a^2}=}$${\displaystyle \frac{8a^3\cdot (a^2-2ax+x^2)}{{(a-x)}^3\cdot 6a^2}}$=

=${\displaystyle \frac{4\cdot \cancel{2}\cancel{a^2}a\cancel{{\left(a-x\right)}^2}}{3\cdot \cancel{2}\cancel{a^2}\cancel{{\left(a-x\right)}^2}(a-x)}=}$${\displaystyle \frac{4a}{3\left(a-x\right)}}$.