618 Вправа 618Алгебра 8 клас Бевз 2025 · § 13 · стор. 125

618. Обчисли: а) √0, √1, √4, √16, √400, √90000; б) √0,01, √0,04, √0,09, √0,16, √0,0081; в) √(1/4), √(1/9), √(4/9), √(25/36), √(16/9).

За означенням: ${\displaystyle \sqrt{a}=}$${\displaystyle b}$, якщо ${\displaystyle b^2=}$${\displaystyle a}$ і ${\displaystyle b\ge 0}$.

а)

${\displaystyle \sqrt{0}=}$${\displaystyle 0}$, бо ${\displaystyle 0^2=}$${\displaystyle 0}$;

${\displaystyle \sqrt{1}=}$${\displaystyle 1}$, бо ${\displaystyle 1^2=}$${\displaystyle 1}$;

${\displaystyle \sqrt{4}=}$${\displaystyle 2}$, бо ${\displaystyle 2^2=}$${\displaystyle 4}$;

${\displaystyle \sqrt{16}=}$${\displaystyle 4}$, бо ${\displaystyle 4^2=}$${\displaystyle 16}$;

${\displaystyle \sqrt{400}=}$${\displaystyle 20}$, бо ${\displaystyle 20^2=}$${\displaystyle 400}$;

${\displaystyle \sqrt{90000}=}$${\displaystyle 300}$, бо ${\displaystyle 300^2=}$${\displaystyle 90000}$;

б)

${\displaystyle \sqrt{0,01}=}$${\displaystyle 0,1}$, бо ${\displaystyle 0,1^2=}$${\displaystyle 0,01}$;

${\displaystyle \sqrt{0,04}=}$${\displaystyle 0,2}$, бо ${\displaystyle 0,2^2=}$${\displaystyle 0,04}$;

${\displaystyle \sqrt{0,09}=}$${\displaystyle 0,3}$, бо ${\displaystyle 0,3^2=}$${\displaystyle 0,09}$;

${\displaystyle \sqrt{0,16}=}$${\displaystyle 0,4}$, бо ${\displaystyle 0,4^2=}$${\displaystyle 0,16}$;

${\displaystyle \sqrt{0,0081}=}$${\displaystyle 0,09}$, бо ${\displaystyle 0,09^2=}$${\displaystyle 0,0081}$;

в)

${\displaystyle \sqrt{\frac{1}{4}}=}$${\displaystyle \frac{1}{2}}$, бо ${\displaystyle {\left(\frac{1}{2}\right)}^2=}$${\displaystyle \frac{1}{4}}$;

${\displaystyle \sqrt{\frac{1}{9}}=}$${\displaystyle \frac{1}{3}}$, бо ${\displaystyle {\left(\frac{1}{3}\right)}^2=}$${\displaystyle \frac{1}{9}}$;

${\displaystyle \sqrt{\frac{4}{9}}=}$${\displaystyle \frac{2}{3}}$, бо ${\displaystyle {\left(\frac{2}{3}\right)}^2=}$${\displaystyle \frac{4}{9}}$;

${\displaystyle \sqrt{\frac{25}{36}}=}$${\displaystyle \frac{5}{6}}$, бо ${\displaystyle {\left(\frac{5}{6}\right)}^2=}$${\displaystyle \frac{25}{36}}$;

${\displaystyle \sqrt{\frac{16}{9}}=}$${\displaystyle \frac{4}{3}}$, бо ${\displaystyle {\left(\frac{4}{3}\right)}^2=}$${\displaystyle \frac{16}{9}}$.