179 Вправа 179Алгебра 8 клас Бевз 2025 · § 4 · стор. 39

179. Спрости вираз.

Врахуємо: ${\displaystyle {(2-a)}^2=}$${\displaystyle {(a-2)}^2}$;

${\displaystyle {(1-a)}^3=}$${\displaystyle -{(a-1)}^3}$.

а) ${\displaystyle \frac{a^2+2a}{{(a-2)}^2}-\frac{6a-4}{{(a-2)}^2}=}$${\displaystyle \frac{a^2+2a-(6a-4)}{{(a-2)}^2}}$=

=${\displaystyle \frac{a^2+2a-6a+4}{{(a-2)}^2}=}$${\displaystyle \frac{a^2-4a+4}{{(a-2)}^2}=}$${\displaystyle \frac{\cancel{{\left(a-2\right)}^2}}{\cancel{{\left(a-2\right)}^2}}=}$${\displaystyle 1}$;

б) ${\displaystyle \frac{m^2}{{(m-5)}^2}-\frac{10m-25}{{(m-5)}^2}=}$${\displaystyle \frac{m^2-(10m-25)}{{(m-5)}^2}=}$${\displaystyle \frac{m^2-10m+25}{{(m-5)}^2}}$=

=${\displaystyle \frac{\cancel{{\left(m-5\right)}^2}}{\cancel{{\left(m-5\right)}^2}}=}$${\displaystyle 1}$;

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

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

г) ${\displaystyle \frac{a+2b}{{(a-b)}^3}+\frac{2a-5b}{{(a-b)}^3}=}$${\displaystyle \frac{a+2b+2a-5b}{{(a-b)}^3}=}$${\displaystyle \frac{3a-3b}{{(a-b)}^3}}$=

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