149 Вправа 149Алгебра 8 клас Бевз 2025 · § 4 · стор. 35

149. Подай у вигляді дробу або одночлена вираз.

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

б) ${\displaystyle \frac{3a}{a^2-1}-\frac{2a}{a^2-1}+\frac{1}{a^2-1}}$=

=${\displaystyle \frac{3a}{(a-1)(a+1)}-\frac{2a}{(a-1)(a+1)}+\frac{1}{(a-1)(a+1)}}$=

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

в) ${\displaystyle \frac{3x+y}{x+y+z}+\frac{z-2x}{x+y+z}=}$${\displaystyle \frac{3x+y+z-2x}{x+y+z}=}$${\displaystyle \frac{\cancel{\left(x+y+z\right)}}{\cancel{\left(x+y+z\right)}}=}$${\displaystyle 1}$;

г) ${\displaystyle \frac{2a-b}{a+b-c}-\frac{a-2b+c}{a+b-c}=}$${\displaystyle \frac{2a-b-(a-2b+c)}{a+b-c}}$=

=${\displaystyle \frac{2a-b-a+2b-c}{a+b-c}=}$${\displaystyle \frac{\cancel{\left(a+b-c\right)}}{\cancel{\left(a+b-c\right)}}=}$${\displaystyle 1}$.