574* Вправа 574Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 164

574. Спростіть вираз:

1) ${\displaystyle 14=}$${\displaystyle 9+5=}$${\displaystyle 3^2+{\left(\sqrt{5}\right)}^2}$; ${\displaystyle 6\sqrt{5}=}$${\displaystyle 2\cdot 3\cdot \sqrt{5}}$; ${\displaystyle 3=}$${\displaystyle \sqrt{9}>\sqrt{5}}$, тому ${\displaystyle 3-\sqrt{5}>0}$;

${\displaystyle \sqrt{14-6\sqrt{5}}=}$${\displaystyle \sqrt{3^2-2\cdot 3\cdot \sqrt{5}+{\left(\sqrt{5}\right)}^2}=}$${\displaystyle \sqrt{{(3-\sqrt{5})}^2}=}$${\displaystyle |3-\sqrt{5}|=}$${\displaystyle 3-\sqrt{5}}$;

2) ${\displaystyle 8=}$${\displaystyle 7+1=}$${\displaystyle {\left(\sqrt{7}\right)}^2+1^2}$; ${\displaystyle 2\sqrt{7}=}$${\displaystyle 2\cdot \sqrt{7}\cdot 1}$; ${\displaystyle \sqrt{7}>\sqrt{1}=}$${\displaystyle 1}$, тому ${\displaystyle \sqrt{7}-1>0}$;

${\displaystyle \sqrt{8-2\sqrt{7}}=}$${\displaystyle \sqrt{{\left(\sqrt{7}\right)}^2-2\cdot \sqrt{7}\cdot 1+1^2}=}$${\displaystyle \sqrt{{(\sqrt{7}-1)}^2}=}$${\displaystyle |\sqrt{7}-1|=}$${\displaystyle \sqrt{7}-1}$;

3) ${\displaystyle 3=}$${\displaystyle 2+1=}$${\displaystyle {\left(\sqrt{2}\right)}^2+1^2}$; ${\displaystyle 2\sqrt{2}=}$${\displaystyle 2\cdot \sqrt{2}\cdot 1}$; ${\displaystyle \sqrt{2}+1>0}$;

${\displaystyle \sqrt{2\sqrt{2}+3}=}$${\displaystyle \sqrt{{\left(\sqrt{2}\right)}^2+2\cdot \sqrt{2}\cdot 1+1^2}=}$${\displaystyle \sqrt{{(\sqrt{2}+1)}^2}=}$${\displaystyle |\sqrt{2}+1|=}$${\displaystyle \sqrt{2}+1}$;

4) ${\displaystyle 7=}$${\displaystyle 4+3=}$${\displaystyle 2^2+{\left(\sqrt{3}\right)}^2}$; ${\displaystyle 4\sqrt{3}=}$${\displaystyle 2\cdot 2\cdot \sqrt{3}}$; ${\displaystyle 2=}$${\displaystyle \sqrt{4}>\sqrt{3}}$, тому ${\displaystyle 2-\sqrt{3}>0}$;

${\displaystyle \sqrt{7-4\sqrt{3}}=}$${\displaystyle \sqrt{2^2-2\cdot 2\cdot \sqrt{3}+{\left(\sqrt{3}\right)}^2}=}$${\displaystyle \sqrt{{(2-\sqrt{3})}^2}=}$${\displaystyle |2-\sqrt{3}|=}$${\displaystyle 2-\sqrt{3}}$.