166 Вправа 166Алгебра 8 клас Бевз 2025 · § 4 · стор. 37

166. Знайди суму і різницю дробів.

а) сума: ${\displaystyle \frac{1^{\mathrm{\setminus }c}}{3cx}+\frac{c-x}{3c^{2}x}=}$${\displaystyle \frac{c+c-x}{3c^{2}x}=}$${\displaystyle \frac{2c-x}{3c^{2}x}}$;

різниця: ${\displaystyle \frac{1^{\mathrm{\setminus }c}}{3cx}-\frac{c-x}{3c^{2}x}=}$${\displaystyle \frac{c-(c-x)}{3c^{2}x}=}$${\displaystyle \frac{c-c+x}{3c^{2}x}}$=

=${\displaystyle \frac{\cancel{x}}{3c^2\cancel{x}}=}$${\displaystyle \frac{1}{3c^2}}$;

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

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

різниця: ${\displaystyle \frac{a^2}{{(a-b)}^2}-\frac{a^{\mathrm{\setminus }a-b}}{a-b}=}$${\displaystyle \frac{a^2-a(a-b)}{{(a-b)}^2}}$=

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

в) сума: ${\displaystyle \frac{3^{\mathrm{\setminus }x}}{x-y}+\frac{3y}{x(x-y)}=}$${\displaystyle \frac{3x+3y}{x(x-y)}=}$${\displaystyle \frac{3\left(x+y\right)}{x\left(x-y\right)}}$;

різниця: ${\displaystyle \frac{3^{\mathrm{\setminus }x}}{x-y}-\frac{3y}{x(x-y)}=}$${\displaystyle \frac{3x-3y}{x(x-y)}}$=

=${\displaystyle \frac{3\cancel{\left(x-y\right)}}{x\cancel{\left(x-y\right)}}=}$${\displaystyle \frac{3}{x}}$.