575* Вправа 575Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 164

575. Доведіть, що:

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

${\displaystyle 4-2\sqrt{3}=}$${\displaystyle {\left(\sqrt{3}\right)}^2-2\cdot \sqrt{3}\cdot 1+1^2=}$${\displaystyle {(\sqrt{3}-1)}^2}$; ${\displaystyle \sqrt{3}>1}$, тому ${\displaystyle \sqrt{3}-1>0}$;

${\displaystyle \sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}=}$${\displaystyle \sqrt{{(\sqrt{3}+1)}^2}+\sqrt{{(\sqrt{3}-1)}^2}=}$${\displaystyle |\sqrt{3}+1|+|\sqrt{3}-1|=}$${\displaystyle \sqrt{3}+1+\sqrt{3}-1=}$${\displaystyle 2\sqrt{3}}$, що і треба було довести;

2) ${\displaystyle 6+4\sqrt{2}=}$${\displaystyle 4+4\sqrt{2}+2=}$${\displaystyle 2^2+2\cdot 2\cdot \sqrt{2}+{\left(\sqrt{2}\right)}^2=}$${\displaystyle {(2+\sqrt{2})}^2}$;

${\displaystyle 6-4\sqrt{2}=}$${\displaystyle 2^2-2\cdot 2\cdot \sqrt{2}+{\left(\sqrt{2}\right)}^2=}$${\displaystyle {(2-\sqrt{2})}^2}$; ${\displaystyle 2=}$${\displaystyle \sqrt{4}>\sqrt{2}}$, тому ${\displaystyle 2-\sqrt{2}>0}$;

${\displaystyle \sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}=}$${\displaystyle \sqrt{{(2+\sqrt{2})}^2}+\sqrt{{(2-\sqrt{2})}^2}=}$${\displaystyle |2+\sqrt{2}|+|2-\sqrt{2}|=}$${\displaystyle 2+\sqrt{2}+2-\sqrt{2}=}$${\displaystyle 4}$, що і треба було довести.