283 Вправа 283Алгебра 8 клас Бевз 2025 · § 6 · стор. 59

283. Виконай ділення.

а) ${\displaystyle \frac{2a^3+16}{a^2+3a+9}:\frac{a^2-2a+4}{2a^3-54}=}$${\displaystyle \frac{2a^3+16}{a^2+3a+9}\cdot \frac{2a^3-54}{a^2-2a+4}}$=

=${\displaystyle \frac{(2a^3+16)\cdot (2a^3-54)}{(a^2+3a+9)\cdot (a^2-2a+4)}}$=

=${\displaystyle \frac{4(a+2)\cancel{\left(a^2+3a+9\right)}\cancel{\left(a^2-2a+4\right)}(a-3)}{\cancel{\left(a^2+3a+9\right)}\cancel{\left(a^2-2a+4\right)}}}$=

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

б) ${\displaystyle \frac{64a^3-b^3}{16a^2-8ab+b^2}:\frac{16a^2+4ab+b^2}{b^2-16a^2}}$=

=${\displaystyle \frac{64a^3-b^3}{16a^2-8ab+b^2}\cdot \frac{b^2-16a^2}{16a^2+4ab+b^2}}$=

=${\displaystyle \frac{(64a^3-b^3)\cdot (b^2-16a^2)}{(16a^2-8ab+b^2)\cdot (16a^2+4ab+b^2)}}$=

=${\displaystyle -\frac{\cancel{\left(16a^2+4ab+b^2\right)}(4a+b)\cancel{{\left(4a-b\right)}^2}}{\cancel{\left(16a^2+4ab+b^2\right)}\cancel{{\left(4a-b\right)}^2}}=}$${\displaystyle -\left(4a+b\right)}$;

в) ${\displaystyle \frac{3y-9}{x^2-xy+x-y}:\frac{xy-3x+2y-6}{x^2-y^2}}$;

${\displaystyle x^2-xy+x-y=}$${\displaystyle x(x-y)+(x-y)=}$${\displaystyle (x-y)(x+1)}$;

${\displaystyle xy-3x+2y-6=}$${\displaystyle x(y-3)+2(y-3)}$=

=${\displaystyle (y-3)(x+2)}$;

${\displaystyle \frac{3(y-3)}{(x-y)(x+1)}\cdot \frac{(x-y)(x+y)}{(y-3)(x+2)}}$=

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

г) ${\displaystyle {\left(\frac{a-1}{a+1}\right)}^2:\frac{a^2-2a+1}{a^2+2a+1}=}$${\displaystyle \frac{{(a-1)}^2}{{(a+1)}^2}\cdot \frac{{(a+1)}^2}{{(a-1)}^2}}$=

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