576* Вправа 576Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 165

576. Порівняйте числа:

1) ${\displaystyle 4\sqrt{3}=}$${\displaystyle \sqrt{48}}$, ${\displaystyle 3\sqrt{4}=}$${\displaystyle \sqrt{36}}$; ${\displaystyle \sqrt{48}>\sqrt{36}}$;

отже, ${\displaystyle \sqrt{4\sqrt{3}}>\sqrt{3\sqrt{4}}}$;

2) порівняємо ${\displaystyle 4-\sqrt{5}}$ і ${\displaystyle 3-\sqrt{2}}$, тобто ${\displaystyle 1+\sqrt{2}}$ і ${\displaystyle \sqrt{5}}$:

${\displaystyle {(1+\sqrt{2})}^2=}$${\displaystyle 3+2\sqrt{2}}$; ${\displaystyle {\left(\sqrt{5}\right)}^2=}$${\displaystyle 5=}$${\displaystyle 3+2}$; ${\displaystyle 2\sqrt{2}>2}$;

отже, ${\displaystyle 1+\sqrt{2}>\sqrt{5}}$ і ${\displaystyle 4-\sqrt{5}>3-\sqrt{2}}$;

3) ${\displaystyle {\left(\sqrt{2\sqrt{5}}\right)}^2=}$${\displaystyle 2\sqrt{5}=}$${\displaystyle \sqrt{20}<5}$; ${\displaystyle {(4-\sqrt{3})}^2=}$${\displaystyle 19-8\sqrt{3}=}$${\displaystyle 19-\sqrt{192}>19-14=}$${\displaystyle 5}$;

отже, ${\displaystyle \sqrt{2\sqrt{5}}<4-\sqrt{3}}$;

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

отже, ${\displaystyle 1+\sqrt{2}>\sqrt{2\sqrt{2}}}$.