175 Вправа 175Алгебра 8 клас Бевз 2025 · § 4 · стор. 38

175. Виконай дії.

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

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

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

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

в) ${\displaystyle \frac{b^2+b-1}{2b-3}-\frac{8+b-3b^2}{2b-3}=}$${\displaystyle \frac{b^2+b-1-(8+b-3b^2)}{2b-3}}$=

=${\displaystyle \frac{b^2+b-1-b+3b^2-8}{2b-3}=}$${\displaystyle \frac{4b^2-9}{2b-3}}$=

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

г) ${\displaystyle \frac{2a^2-3}{a^2-9}-\frac{a+1}{a^2-9}+\frac{4+7a}{a^2-9}}$=

=${\displaystyle \frac{2a^2-3}{(a-3)(a+3)}-\frac{a+1}{(a-3)(a+3)}+\frac{4+7a}{(a-3)(a+3)}}$=

=${\displaystyle \frac{2a^2-3-(a+1)+4+7a}{(a-3)(a+3)}=}$${\displaystyle \frac{2a^2-3-a-1+7a+4}{(a-3)(a+3)}}$=

=${\displaystyle \frac{2a^2+6a}{(a-3)(a+3)}=}$${\displaystyle \frac{2a\cancel{\left(a+3\right)}}{(a-3)\cancel{\left(a+3\right)}}=}$${\displaystyle \frac{2a}{a-3}}$.