185 Вправа 185Алгебра 8 клас Бевз 2025 · § 4 · стор. 39

185. Доведи тотожність.

а) ${\displaystyle \frac{{(m+2)}^2}{8m}-\frac{{(m-2)}^2}{8m}=}$${\displaystyle \frac{{(m+2)}^2-{(m-2)}^2}{8m}}$=

=${\displaystyle \frac{m^2+4m+4-m^2+4m-4}{8m}=}$${\displaystyle \frac{\cancel{8m}}{\cancel{8m}}=}$${\displaystyle 1}$;

${\displaystyle 1=}$${\displaystyle 1}$ — тотожність доведено;

б) ${\displaystyle \frac{5m-10}{3m-9}-\frac{m-8}{9-3m}=}$${\displaystyle \frac{5m-10}{3(m-3)}+\frac{m-8}{3(m-3)}}$=

=${\displaystyle \frac{5m-10+m-8}{3(m-3)}=}$${\displaystyle \frac{6m-18}{3(m-3)}=}$${\displaystyle \frac{2\cancel{3\left(m-3\right)}}{\cancel{3\left(m-3\right)}}=}$${\displaystyle 2}$;

${\displaystyle 2=}$${\displaystyle 2}$ — тотожність доведено;

в) ${\displaystyle \frac{a^2-3ab+9b^2}{a-3b}+\frac{a^2+3ab+9b^2}{a+3b}-\frac{18a{b}^2}{a^2-9b^2}}$=

=${\displaystyle \frac{{a^2-3ab+9b^2}^{\mathrm{\setminus }a+3b}}{a-3b}+\frac{{a^2+3ab+9b^2}^{\mathrm{\setminus }a-3b}}{a+3b}-\frac{18a{b}^2}{(a-3b)(a+3b)}}$=

=${\displaystyle \frac{(a^2-3ab+9b^2)(a+3b)+(a^2+3ab+9b^2)(a-3b)-18a{b}^2}{(a-3b)(a+3b)}}$=

=${\displaystyle \frac{a^3+27b^3+a^3-27b^3-18a{b}^2}{(a-3b)(a+3b)}=}$${\displaystyle \frac{2a^3-18a{b}^2}{(a-3b)(a+3b)}}$=

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

${\displaystyle 2a=}$${\displaystyle 2a}$ — тотожність доведено.