244 Вправа 244Алгебра 8 клас Бевз 2025 · § 5 · стор. 53

244. Спрости вираз.

а) ${\displaystyle (-8x^6{y}^3)\cdot {\left(\frac{3z^4}{2x^{2}y}\right)}^3=}$${\displaystyle -8x^6{y}^3\cdot \frac{27z^{12}}{8x^6{y}^3}}$=

=${\displaystyle -\frac{\cancel{8}\cancel{x^6}\cancel{y^3}\cdot 27z^{12}}{\cancel{8}\cancel{x^6}\cancel{y^3}}=}$${\displaystyle -27z^{12}}$;

б) ${\displaystyle {\left(\frac{a^2}{3b^3}\right)}^3\cdot {\left(\frac{9b^4}{a^3}\right)}^2=}$${\displaystyle \frac{a^6}{27b^9}\cdot \frac{81b^8}{a^6}=}$${\displaystyle \frac{a^6\cdot 81b^8}{27b^9\cdot a^6}}$=

=${\displaystyle \frac{\cancel{a^6}\cdot 3\cdot \cancel{27}\cancel{b^8}}{\cancel{27}b\cancel{b^8}\cdot \cancel{a^6}}=}$${\displaystyle \frac{3}{b}}$;

в) ${\displaystyle \frac{x^2-4x+4}{x^2-4}\cdot {\left(\frac{x+2}{x-2}\right)}^2=}$${\displaystyle \frac{{(x-2)}^2}{(x-2)(x+2)}\cdot \frac{{(x+2)}^2}{{(x-2)}^2}}$=

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

г) ${\displaystyle \frac{a-c}{a^2-ac+c^2}\cdot \frac{a^3+c^3}{a^2-c^2}=}$${\displaystyle \frac{(a-c)\cdot (a^3+c^3)}{(a^2-ac+c^2)\cdot (a^2-c^2)}}$=

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