284 Вправа 284Алгебра 8 клас Бевз 2025 · § 6 · стор. 60

284. Виконай ділення.

а) ${\displaystyle \frac{a^2-2ab+a-2b}{a{b}^2+a^{2}b}:\frac{a^2-4b^2}{a^2+ab+a+b}}$;

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

=${\displaystyle (a-2b)(a+1)}$;

${\displaystyle a^2+ab+a+b=}$${\displaystyle a(a+b)+(a+b)=}$${\displaystyle (a+b)(a+1)}$;

${\displaystyle \frac{(a-2b)(a+1)}{ab(b+a)}\cdot \frac{(a+b)(a+1)}{(a-2b)(a+2b)}}$=

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

б) ${\displaystyle {\left(\frac{x+z}{2x}\right)}^2:\frac{x^2+z^2+2xz}{8x^4}=}$${\displaystyle \frac{{(x+z)}^2}{4x^2}\cdot \frac{8x^4}{x^2+z^2+2xz}}$=

=${\displaystyle \frac{{(x+z)}^2\cdot 8x^4}{4x^2\cdot (x^2+z^2+2xz)}=}$${\displaystyle \frac{2\cdot \cancel{4}\cancel{x^2}{x}^2\cancel{{\left(x+z\right)}^2}}{\cancel{4}\cancel{x^2}\cancel{{\left(x+z\right)}^2}}=}$${\displaystyle 2x^2}$.