828 Вправа 828Алгебра 8 клас Бевз 2025 · § 16 · стор. 156

828. Скороти дріб.

Розкладаємо чисельник і знаменник на множники та скорочуємо спільний множник.

а) ${\displaystyle \frac{\sqrt{a}-1}{a-1}=}$${\displaystyle \frac{\cancel{\sqrt{a}-1}}{\cancel{\left(\sqrt{a}-1\right)}(\sqrt{a}+1)}=}$${\displaystyle \frac{1}{\sqrt{a}+1}}$;

б) ${\displaystyle \frac{\sqrt{x}-\sqrt{z}}{x-z}=}$${\displaystyle \frac{\cancel{\sqrt{x}-\sqrt{z}}}{\cancel{\left(\sqrt{x}-\sqrt{z}\right)}(\sqrt{x}+\sqrt{z})}=}$${\displaystyle \frac{1}{\sqrt{x}+\sqrt{z}}}$;

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

г) ${\displaystyle \frac{\sqrt{a}+1}{\sqrt{a}+a}=}$${\displaystyle \frac{\cancel{\sqrt{a}+1}}{\sqrt{a}\cancel{\left(1+\sqrt{a}\right)}}=}$${\displaystyle \frac{1}{\sqrt{a}}}$;

ґ) ${\displaystyle \frac{\sqrt{a}+\sqrt{x}}{a\sqrt{x}+x\sqrt{a}}=}$${\displaystyle \frac{\cancel{\sqrt{a}+\sqrt{x}}}{\sqrt{ax}\cancel{\left(\sqrt{a}+\sqrt{x}\right)}}=}$${\displaystyle \frac{1}{\sqrt{ax}}}$;

д) ${\displaystyle \frac{a+2\sqrt{a}+1}{a+\sqrt{a}}=}$${\displaystyle \frac{{(\sqrt{a}+1)}^2}{\sqrt{a}(\sqrt{a}+1)}=}$${\displaystyle \frac{\sqrt{a}+1}{\sqrt{a}}}$;

е) ${\displaystyle \frac{x-6\sqrt{x}+9}{x-9}=}$${\displaystyle \frac{{(\sqrt{x}-3)}^2}{(\sqrt{x}-3)(\sqrt{x}+3)}=}$${\displaystyle \frac{\sqrt{x}-3}{\sqrt{x}+3}}$;

є) ${\displaystyle \frac{{(\sqrt{3}+1)}^2}{2+\sqrt{3}}=}$${\displaystyle \frac{4+2\sqrt{3}}{2+\sqrt{3}}=}$${\displaystyle \frac{2\cancel{\left(2+\sqrt{3}\right)}}{\cancel{2+\sqrt{3}}}=}$${\displaystyle 2}$;

ж) ${\displaystyle \frac{{(\sqrt{7}-1)}^2}{\sqrt{7}-4}=}$${\displaystyle \frac{8-2\sqrt{7}}{\sqrt{7}-4}=}$${\displaystyle \frac{-2\cancel{\left(\sqrt{7}-4\right)}}{\cancel{\sqrt{7}-4}}=}$${\displaystyle -2}$.