98 Вправа 98Алгебра 8 клас Бевз 2025 · § 2 · стор. 23

98. Спрости вираз.

а) ${\displaystyle \frac{ac+bx+ax+bc}{ay+2bx+2ax+by}}$;

${\displaystyle \frac{(ac+bc)+(ax+bx)}{(ay+by)+(2ax+2bx)}=}$${\displaystyle \frac{c(a+b)+x(a+b)}{y(a+b)+2x(a+b)}}$=

=${\displaystyle \frac{\cancel{\left(a+b\right)}(c+x)}{\cancel{\left(a+b\right)}(y+2x)}=}$${\displaystyle \frac{c+x}{y+2x}}$;

б) ${\displaystyle \frac{x^2-(a-b)x-ab}{x^3+b{x}^2+ax+ab}}$;

${\displaystyle \frac{x^2-ax+bx-ab}{x^2(x+b)+a(x+b)}=}$${\displaystyle \frac{x(x-a)+b(x-a)}{(x+b)(x^2+a)}}$=

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

в) ${\displaystyle \frac{{(x+a)}^2-{(z+c)}^2}{{(x+z)}^2-{(a+c)}^2}}$;

${\displaystyle \frac{(x+a-z-c)\cancel{\left(x+a+z+c\right)}}{(x+z-a-c)\cancel{\left(x+z+a+c\right)}}=}$${\displaystyle \frac{x+a-z-c}{x+z-a-c}}$.