248 Вправа 248Алгебра 8 клас Бевз 2025 · § 5 · стор. 53

248. Спрости вираз.

а) ${\displaystyle -8a^2(a+b)\cdot \frac{a^2+b^2}{a^2-b^2}\cdot \frac{b-a}{{\left(8a\right)}^2}}$=

=${\displaystyle \frac{-8a^2(a+b)\cdot (a^2+b^2)\cdot (b-a)}{(a^2-b^2)\cdot {\left(8a\right)}^2}}$=

=${\displaystyle \frac{\cancel{8}\cancel{a^2}\cancel{\left(a+b\right)}(a^2+b^2)\cancel{\left(a-b\right)}}{8\cdot \cancel{8}\cancel{a^2}\cancel{\left(a+b\right)}\cancel{\left(a-b\right)}}=}$${\displaystyle \frac{a^2+b^2}{8}}$;

б) ${\displaystyle \frac{x^2-x}{2x+2}\cdot \frac{x^2+2x+1}{x^2+4x}\cdot \frac{x^2-16}{3x-3}}$=

=${\displaystyle \frac{(x^2-x)\cdot (x^2+2x+1)\cdot (x^2-16)}{(2x+2)\cdot (x^2+4x)\cdot (3x-3)}}$=

=${\displaystyle \frac{\cancel{x}\cancel{\left(x+1\right)}(x+1)\cancel{\left(x+4\right)}\cancel{\left(x-1\right)}(x-4)}{6\cancel{x}\cancel{\left(x+1\right)}\cancel{\left(x+4\right)}\cancel{\left(x-1\right)}}=}$${\displaystyle \frac{\left(x-4\right)\left(x+1\right)}{6}}$.