833 Вправа 833Алгебра 8 клас Бевз 2025 · § 16 · стор. 156

833. Звільнися від ірраціональності у знаменнику дробу.

а) ${\displaystyle \frac{\sqrt{2}-1}{\sqrt{2}+1}=}$${\displaystyle \frac{(\sqrt{2}-1)\left(\sqrt{2}-1\right)}{2-1}}$=

=${\displaystyle 2-2\sqrt{2}+1=}$${\displaystyle 3-2\sqrt{2}}$;

б) ${\displaystyle \frac{a-1}{\sqrt{a+3}-2}=}$${\displaystyle \frac{(a-1)\left(\sqrt{a+3}+2\right)}{a+3-4}}$=

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

в) ${\displaystyle \frac{\sqrt{12}-\sqrt{8}}{\sqrt{3}+\sqrt{2}}=}$${\displaystyle \frac{2(\sqrt{3}-\sqrt{2})\left(\sqrt{3}-\sqrt{2}\right)}{3-2}}$=

=${\displaystyle 2(5-2\sqrt{6})=}$${\displaystyle 10-4\sqrt{6}}$;

г) ${\displaystyle \frac{x}{\sqrt{1-x}-\sqrt{1-2x}}=}$${\displaystyle \frac{x\left(\sqrt{1-x}+\sqrt{1-2x}\right)}{1-x-1+2x}}$=

=${\displaystyle \frac{\cancel{x}(\sqrt{1-x}+\sqrt{1-2x})}{\cancel{x}}=}$${\displaystyle \sqrt{1-x}+\sqrt{1-2x}}$.