170 Вправа 170Алгебра 8 клас Бевз 2025 · § 4 · стор. 37

170. Спрости вираз.

а) ${\displaystyle \frac{6}{a^2+2a}-\frac{3}{a}=}$${\displaystyle \frac{6}{a(a+2)}-\frac{3^{\mathrm{\setminus }a+2}}{a}=}$${\displaystyle \frac{6-3(a+2)}{a(a+2)}}$=

=${\displaystyle \frac{6-3a-6}{a(a+2)}=}$${\displaystyle \frac{-3\cancel{a}}{(a+2)\cancel{a}}=}$${\displaystyle -\frac{3}{a+2}}$;

б) ${\displaystyle \frac{3a-1}{a^2-a}-\frac{2}{a-1}=}$${\displaystyle \frac{3a-1}{a(a-1)}-\frac{2^{\mathrm{\setminus }a}}{a-1}=}$${\displaystyle \frac{3a-1-2a}{a(a-1)}}$=

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

в) ${\displaystyle \frac{2c-1}{3c-3}-\frac{3c-1}{6c-6}=}$${\displaystyle \frac{{2c-1}^{\mathrm{\setminus }2}}{3(c-1)}-\frac{3c-1}{6(c-1)}}$=

=${\displaystyle \frac{2(2c-1)-(3c-1)}{6(c-1)}=}$${\displaystyle \frac{4c-2-3c+1}{6(c-1)}=}$${\displaystyle \frac{\cancel{\left(c-1\right)}}{6\cancel{\left(c-1\right)}}=}$${\displaystyle \frac{1}{6}}$;

г) ${\displaystyle \frac{2a-b}{a^2-ab}+\frac{a-2b}{ab-b^2}=}$${\displaystyle \frac{{2a-b}^{\mathrm{\setminus }b}}{a(a-b)}+\frac{{a-2b}^{\mathrm{\setminus }a}}{b(a-b)}}$=

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

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

ґ) ${\displaystyle \frac{a}{ab-b^2}+\frac{b}{ab-a^2}=}$${\displaystyle \frac{a^{\mathrm{\setminus }a}}{b(a-b)}-\frac{b^{\mathrm{\setminus }b}}{a(a-b)}=}$${\displaystyle \frac{a^2-b^2}{ab(a-b)}}$=

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

д) ${\displaystyle \frac{c+9}{9c-81}-\frac{9-3c}{9c-c^2}=}$${\displaystyle \frac{{c+9}^{\mathrm{\setminus }c}}{9(c-9)}+\frac{{9-3c}^{\mathrm{\setminus }9}}{c(c-9)}}$=

=${\displaystyle \frac{c(c+9)+9(9-3c)}{9c(c-9)}=}$${\displaystyle \frac{c^2+9c+81-27c}{9c(c-9)}=}$${\displaystyle \frac{c^2-18c+81}{9c(c-9)}}$=

=${\displaystyle \frac{(c-9)\cancel{\left(c-9\right)}}{9c\cancel{\left(c-9\right)}}=}$${\displaystyle \frac{c-9}{9c}}$.