249 Вправа 249Алгебра 8 клас Бевз 2025 · § 5 · стор. 53

249. Знайди значення виразу.

а) ${\displaystyle \frac{a{x}^2-a}{x^2+ax+a^2}\cdot \frac{a^3-x^3}{4+4x}=}$${\displaystyle \frac{(a{x}^2-a)\cdot (a^3-x^3)}{(x^2+ax+a^2)\cdot (4+4x)}}$=

=${\displaystyle \frac{a\cancel{\left(a^2+ax+x^2\right)}(a-x)\cancel{\left(x+1\right)}(x-1)}{4\cancel{\left(a^2+ax+x^2\right)}\cancel{\left(x+1\right)}}=}$${\displaystyle \frac{a\left(x-1\right)\left(a-x\right)}{4}}$;

якщо a = 4, x = 3, то ${\displaystyle \frac{4\cdot (3-1)(4-3)}{4}}$=

=${\displaystyle \frac{4\cdot 2\cdot 1}{4}=}$${\displaystyle 2}$;

б) ${\displaystyle \frac{ab-2b}{{(b-3)}^2-4a^2}\cdot \frac{4a^2+2ab+3b-9}{a^2-2a}}$;

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

${\displaystyle 4a^2+2ab+3b-9=}$${\displaystyle (2a-3)(2a+3)+b(2a+3)}$=

=${\displaystyle (2a+3)(2a+b-3)}$;

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

якщо a = 1/2, b = 3, то ${\displaystyle \frac{3\cdot (1+3)}{\frac{1}{2}\cdot (3-3-1)}}$=

=${\displaystyle \frac{12}{-\frac{1}{2}}=}$${\displaystyle -24}$.