562 Вправа 562Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 163

562. Спростіть вираз:

1) ${\displaystyle 5=}$${\displaystyle \sqrt{25}>\sqrt{23}}$, тому ${\displaystyle 5-\sqrt{23}>0}$;

${\displaystyle \sqrt{{(5-\sqrt{23})}^2}=}$${\displaystyle |5-\sqrt{23}|=}$${\displaystyle 5-\sqrt{23}}$;

2) ${\displaystyle 2=}$${\displaystyle \sqrt{4}>\sqrt{3}}$, тому ${\displaystyle \sqrt{3}-2<0}$;

${\displaystyle \sqrt{{(\sqrt{3}-2)}^2}=}$${\displaystyle |\sqrt{3}-2|=}$${\displaystyle -(\sqrt{3}-2)=}$${\displaystyle 2-\sqrt{3}}$;

3) ${\displaystyle \sqrt{8}<\sqrt{12}}$, тому ${\displaystyle \sqrt{8}-\sqrt{12}<0}$;

${\displaystyle \sqrt{{(\sqrt{8}-\sqrt{12})}^2}=}$${\displaystyle |\sqrt{8}-\sqrt{12}|=}$${\displaystyle -(\sqrt{8}-\sqrt{12})=}$${\displaystyle \sqrt{12}-\sqrt{8}=}$${\displaystyle 2\sqrt{3}-2\sqrt{2}}$;

4) ${\displaystyle 1=}$${\displaystyle \sqrt{1}<\sqrt{2}}$, тому ${\displaystyle 1-\sqrt{2}<0}$;

${\displaystyle \sqrt{{(1-\sqrt{2})}^2}+\sqrt{{(1+\sqrt{2})}^2}=}$${\displaystyle |1-\sqrt{2}|+|1+\sqrt{2}|=}$${\displaystyle -(1-\sqrt{2})+1+\sqrt{2}=}$${\displaystyle \sqrt{2}-1+1+\sqrt{2}=}$${\displaystyle 2\sqrt{2}}$.