286 Вправа 286Алгебра 8 клас Бевз 2025 · § 6 · стор. 60

286. Знайди значення виразу.

а) ${\displaystyle \frac{ab+4b-5a-20}{b^2-1}:\frac{2b-10}{b+1}}$;

${\displaystyle ab+4b-5a-20=}$${\displaystyle b(a+4)-5(a+4)}$=

=${\displaystyle (a+4)(b-5)}$;

${\displaystyle \frac{(a+4)(b-5)}{(b-1)(b+1)}\cdot \frac{b+1}{2(b-5)}=}$${\displaystyle \frac{(a+4)\cancel{\left(b-5\right)}\cancel{\left(b+1\right)}}{(b-1)\cancel{\left(b+1\right)}\cdot 2\cancel{\left(b-5\right)}}=}$${\displaystyle \frac{a+4}{2(b-1)}}$;

якщо a = 4, b = 3/4, то ${\displaystyle \frac{4+4}{2\cdot (\frac{3}{4}-1)}}$=

=${\displaystyle \frac{8}{-\frac{1}{2}}=}$${\displaystyle -16}$;

б) ${\displaystyle \frac{9-{(a+x)}^2}{6a+6x}:\frac{a^2+3a-x^2-3x}{a^2+ax}}$;

${\displaystyle 9-{(a+x)}^2=}$${\displaystyle (3-a-x)(3+a+x)}$;

${\displaystyle a^2-x^2+3a-3x=}$${\displaystyle (a-x)(a+x)+3(a-x)}$=

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

${\displaystyle \frac{(3-a-x)(3+a+x)}{6(a+x)}\cdot \frac{a(a+x)}{(a-x)(a+x+3)}}$=

=${\displaystyle \frac{(3-a-x)\cancel{\left(3+a+x\right)}a\cancel{\left(a+x\right)}}{6\cancel{\left(a+x\right)}(a-x)\cancel{\left(a+x+3\right)}}=}$${\displaystyle \frac{a(3-a-x)}{6(a-x)}}$;

якщо a = 3/2, x = 1/2, то ${\displaystyle \frac{\frac{3}{2}\cdot (3-\frac{3}{2}-\frac{1}{2})}{6\cdot (\frac{3}{2}-\frac{1}{2})}}$=

=${\displaystyle \frac{\frac{3}{2}\cdot 1}{6\cdot 1}=}$${\displaystyle \frac{1}{4}}$.