176 Вправа 176Алгебра 8 клас Бевз 2025 · § 4 · стор. 38

176. Виконай дії.

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

б) ${\displaystyle \frac{b^2}{{(b-3)}^2}-\frac{6b-9}{{(b-3)}^2}=}$${\displaystyle \frac{b^2-(6b-9)}{{(b-3)}^2}=}$${\displaystyle \frac{b^2-6b+9}{{(b-3)}^2}}$=

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

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

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

=${\displaystyle \frac{3a^2+b^2-2b-2b^2+a^2+2b}{2a-b}=}$${\displaystyle \frac{4a^2-b^2}{2a-b}}$=

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

г) ${\displaystyle \frac{4a^2+5}{a^2-4}-\frac{a^2-2a+8}{a^2-4}+\frac{4a+3}{a^2-4}}$=

=${\displaystyle \frac{4a^2+5}{(a-2)(a+2)}-\frac{a^2-2a+8}{(a-2)(a+2)}+\frac{4a+3}{(a-2)(a+2)}}$=

=${\displaystyle \frac{4a^2+5-(a^2-2a+8)+4a+3}{(a-2)(a+2)}}$=

=${\displaystyle \frac{4a^2+5-a^2+2a-8+4a+3}{(a-2)(a+2)}=}$${\displaystyle \frac{3a^2+6a}{(a-2)(a+2)}}$=

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