153 Вправа 153Алгебра 8 клас Бевз 2025 · § 4 · стор. 36

153. Подай у вигляді дробу або одночлена вираз.

а) ${\displaystyle \frac{5x}{2y}-\frac{x+y}{2y}+\frac{2x+y}{2y}=}$${\displaystyle \frac{5x-(x+y)+2x+y}{2y}}$=

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

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

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

в) ${\displaystyle \frac{2a-5}{2a-3}+\frac{a-2}{2a-3}-\frac{a-4}{2a-3}=}$${\displaystyle \frac{2a-5+a-2-(a-4)}{2a-3}}$=

=${\displaystyle \frac{2a-5+a-2-a+4}{2a-3}=}$${\displaystyle \frac{\cancel{\left(2a-3\right)}}{\cancel{\left(2a-3\right)}}=}$${\displaystyle 1}$;

г) ${\displaystyle \frac{x^3}{x^2-4}-\frac{x^3-x}{x^2-4}-\frac{2}{x^2-4}}$=

=${\displaystyle \frac{x^3}{(x-2)(x+2)}-\frac{x^3-x}{(x-2)(x+2)}-\frac{2}{(x-2)(x+2)}}$=

=${\displaystyle \frac{x^3-(x^3-x)-2}{(x-2)(x+2)}=}$${\displaystyle \frac{x^3-x^3+x-2}{(x-2)(x+2)}=}$${\displaystyle \frac{\cancel{\left(x-2\right)}}{(x+2)\cancel{\left(x-2\right)}}=}$${\displaystyle \frac{1}{x+2}}$.