99 Вправа 99Алгебра 8 клас Бевз 2025 · § 2 · стор. 23

99. Спрости вираз.

а) ${\displaystyle \frac{x-xy+z-zy}{1-3y+3y^2-y^3}}$;

${\displaystyle 1-3y+3y^2-y^3=}$${\displaystyle {(1-y)}^3}$ — куб різниці;

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

б) ${\displaystyle \frac{x^4+(2c^2-a^2)x^2+c^4}{x^4+2a{x}^3+a^2{x}^2-c^4}}$;

чисельник: ${\displaystyle x^4+2c^2{x}^2+c^4-a^2{x}^2}$=

=${\displaystyle {(x^2+c^2)}^2-{\left(ax\right)}^2}$=

=${\displaystyle (x^2-ax+c^2)(x^2+ax+c^2)}$;

знаменник: ${\displaystyle {(x^2+ax)}^2-{\left(c^2\right)}^2}$=

=${\displaystyle (x^2+ax-c^2)(x^2+ax+c^2)}$;

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

в) ${\displaystyle \frac{a^{3}c-2a^2{c}^2+a{c}^3-a{b}^{2}c}{{(a^2+c^2-b^2)}^2-4a^2{c}^2}}$;

чисельник: ${\displaystyle ac(a^2-2ac+c^2-b^2)}$=

=${\displaystyle ac({(a-c)}^2-b^2)=}$${\displaystyle ac(a-c-b)(a-c+b)}$;

знаменник: ${\displaystyle (a^2+c^2-b^2-2ac)(a^2+c^2-b^2+2ac)}$=

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

=${\displaystyle (a-c-b)(a-c+b)(a+c-b)(a+c+b)}$;

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