243 Вправа 243Алгебра 8 клас Бевз 2025 · § 5 · стор. 52

243. Спрости вираз.

а) ${\displaystyle {(-\frac{3c^2}{2a^2{b}^3})}^2\cdot 12a^4{b}^6=}$${\displaystyle \frac{9c^4}{4a^4{b}^6}\cdot 12a^4{b}^6}$=

=${\displaystyle \frac{9c^4\cdot 12a^4{b}^6}{4a^4{b}^6}=}$${\displaystyle \frac{9c^4\cdot 3\cdot \cancel{4}\cancel{a^4}\cancel{b^6}}{\cancel{4}\cancel{a^4}\cancel{b^6}}=}$${\displaystyle 27c^4}$;

б) ${\displaystyle {\left(\frac{x^2}{-2y^3}\right)}^2\cdot {\left(\frac{-2y^2}{x^3}\right)}^3=}$${\displaystyle \frac{x^4}{4y^6}\cdot \frac{-8y^6}{x^9}=}$${\displaystyle \frac{x^4\cdot (-8y^6)}{4y^6\cdot x^9}}$=

=${\displaystyle -\frac{\cancel{x^4}\cdot 2\cdot \cancel{4}\cancel{y^6}}{\cancel{4}\cancel{y^6}\cdot x^5\cancel{x^4}}=}$${\displaystyle -\frac{2}{x^5}}$;

в) ${\displaystyle \frac{a{x}^3}{5-5a}\cdot \frac{a^2-2a+1}{0,2a{x}^3}=}$${\displaystyle \frac{a{x}^3\cdot {(a-1)}^2}{5(1-a)\cdot 0,2a{x}^3}}$=

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

=${\displaystyle -(a-1)=}$${\displaystyle 1-a}$;

г) ${\displaystyle {\left(\frac{2a-1}{a-1}\right)}^2\cdot \frac{a^3-1}{4a^2-4a+1}=}$${\displaystyle \frac{{(2a-1)}^2}{{(a-1)}^2}\cdot \frac{(a-1)(a^2+a+1)}{{(2a-1)}^2}}$=

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