770 Вправа 770Алгебра 8 клас Бевз 2025 · § 15 · стор. 146

770. Обчисли.

${\displaystyle \sqrt{17}\approx 4,12}$, ${\displaystyle \sqrt{11}\approx 3,32}$.

а) ${\displaystyle \sqrt{17}-3>0}$, ${\displaystyle 12-\sqrt{17}>0}$;

${\displaystyle \sqrt{{(\sqrt{17}-3)}^2}+\sqrt{{(12-\sqrt{17})}^2}}$=

=${\displaystyle |\sqrt{17}-3|+|12-\sqrt{17}|}$=

=${\displaystyle \sqrt{17}-3+12-\sqrt{17}=}$${\displaystyle 9}$;

б) ${\displaystyle \sqrt{11}-4<0}$, ${\displaystyle 3-\sqrt{11}<0}$;

${\displaystyle \sqrt{{(\sqrt{11}-4)}^2}+\sqrt{{(3-\sqrt{11})}^2}}$=

=${\displaystyle |\sqrt{11}-4|+|3-\sqrt{11}|}$=

=${\displaystyle 4-\sqrt{11}+\sqrt{11}-3=}$${\displaystyle 1}$.