561 Вправа 561Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 162

561. Спростіть вираз:

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

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

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

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

3) ${\displaystyle 3=}$${\displaystyle \sqrt{9}>\sqrt{5}}$, тому ${\displaystyle 3-\sqrt{5}>0}$;

${\displaystyle \sqrt{{(3-\sqrt{5})}^2}=}$${\displaystyle |3-\sqrt{5}|=}$${\displaystyle 3-\sqrt{5}}$;

4) ${\displaystyle 3=}$${\displaystyle \sqrt{9}<\sqrt{10}}$, тому ${\displaystyle 3-\sqrt{10}<0}$;

${\displaystyle \sqrt{{(3-\sqrt{10})}^2}=}$${\displaystyle |3-\sqrt{10}|=}$${\displaystyle -(3-\sqrt{10})=}$${\displaystyle \sqrt{10}-3}$;

5) ${\displaystyle \sqrt{2}<\sqrt{3}}$, тому ${\displaystyle \sqrt{2}-\sqrt{3}<0}$;

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

6) ${\displaystyle \sqrt{7}>\sqrt{6}}$, тому ${\displaystyle \sqrt{7}-\sqrt{6}>0}$;

${\displaystyle \sqrt{{(\sqrt{7}-\sqrt{6})}^2}=}$${\displaystyle |\sqrt{7}-\sqrt{6}|=}$${\displaystyle \sqrt{7}-\sqrt{6}}$;

7) ${\displaystyle 2\sqrt{3}=}$${\displaystyle \sqrt{12}<\sqrt{15}}$, тому ${\displaystyle \sqrt{15}-2\sqrt{3}>0}$;

${\displaystyle \sqrt{{(\sqrt{15}-2\sqrt{3})}^2}=}$${\displaystyle |\sqrt{15}-2\sqrt{3}|=}$${\displaystyle \sqrt{15}-2\sqrt{3}}$;

8) ${\displaystyle 3\sqrt{3}=}$${\displaystyle \sqrt{27}>\sqrt{25}=}$${\displaystyle 5}$, тому ${\displaystyle 3\sqrt{3}-5>0}$;

${\displaystyle \sqrt{{(3\sqrt{3}-5)}^2}=}$${\displaystyle |3\sqrt{3}-5|=}$${\displaystyle 3\sqrt{3}-5}$;

9) ${\displaystyle 1=}$${\displaystyle \sqrt{1}<\sqrt{8}}$, тому ${\displaystyle 1-\sqrt{8}<0}$;

${\displaystyle \sqrt{{(1-\sqrt{8})}^2}+\sqrt{{(1+\sqrt{8})}^2}=}$${\displaystyle |1-\sqrt{8}|+|1+\sqrt{8}|=}$${\displaystyle -(1-\sqrt{8})+1+\sqrt{8}=}$${\displaystyle \sqrt{8}-1+1+\sqrt{8}=}$${\displaystyle 2\sqrt{8}=}$${\displaystyle 2\sqrt{4\cdot 2}=}$${\displaystyle 4\sqrt{2}}$;

10) ${\displaystyle 1=}$${\displaystyle \sqrt{1}<\sqrt{7}}$, тому ${\displaystyle 1-\sqrt{7}<0}$;

${\displaystyle \sqrt{{(1-\sqrt{7})}^2}-\sqrt{{(3+\sqrt{7})}^2}=}$${\displaystyle |1-\sqrt{7}|-|3+\sqrt{7}|=}$${\displaystyle -(1-\sqrt{7})-(3+\sqrt{7})=}$${\displaystyle \sqrt{7}-1-3-\sqrt{7}=}$${\displaystyle -4}$.