161 Вправа 161Алгебра 8 клас Бевз 2025 · § 4 · стор. 37

161. Додай дроби.

а) ${\displaystyle \frac{1^{\mathrm{\setminus }2}}{x}+\frac{3}{2x}=}$${\displaystyle \frac{2+3}{2x}=}$${\displaystyle \frac{5}{2x}}$;

б) ${\displaystyle \frac{a^{\mathrm{\setminus }4}}{c}+\frac{3-a}{4c}=}$${\displaystyle \frac{4a+3-a}{4c}=}$${\displaystyle \frac{3a+3}{4c}=}$${\displaystyle \frac{3\left(a+1\right)}{4c}}$;

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

=${\displaystyle \frac{6a+3x+2x-6a}{6x}=}$${\displaystyle \frac{5\cancel{x}}{6\cancel{x}}=}$${\displaystyle \frac{5}{6}}$;

г) ${\displaystyle \frac{1^{\mathrm{\setminus }x}}{a+b}+\frac{3}{x(a+b)}=}$${\displaystyle \frac{x+3}{x\left(a+b\right)}}$;

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

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

д) ${\displaystyle \frac{1^{\mathrm{\setminus }3x}}{2a^{2}b{x}^3}+\frac{5^{\mathrm{\setminus }2a}}{3ab{x}^4}=}$${\displaystyle \frac{3x+10a}{6a^{2}b{x}^4}}$.