670 Вправа 670Алгебра 8 клас Бевз 2025 · § 13 · стор. 130

670. Розв’яжи рівняння.

а) ${\displaystyle \sqrt{x+2}=}$${\displaystyle 3}$;

${\displaystyle x+2=}$${\displaystyle 3^2}$;

${\displaystyle x+2=}$${\displaystyle 9}$;

${\displaystyle x=}$${\displaystyle 9-2}$;

${\displaystyle x=}$${\displaystyle 7}$;

б) ${\displaystyle \sqrt{x-12}=}$${\displaystyle 8}$;

${\displaystyle x-12=}$${\displaystyle 8^2}$;

${\displaystyle x-12=}$${\displaystyle 64}$;

${\displaystyle x=}$${\displaystyle 64+12}$;

${\displaystyle x=}$${\displaystyle 76}$;

в) ${\displaystyle \frac{36}{\sqrt{x-5}}=}$${\displaystyle 4}$;

${\displaystyle \sqrt{x-5}=}$${\displaystyle 36:4}$;

${\displaystyle \sqrt{x-5}=}$${\displaystyle 9}$;

${\displaystyle x-5=}$${\displaystyle 9^2}$;

${\displaystyle x-5=}$${\displaystyle 81}$;

${\displaystyle x=}$${\displaystyle 81+5}$;

${\displaystyle x=}$${\displaystyle 86}$;

г) ${\displaystyle \frac{15}{\sqrt{x}-3}=}$${\displaystyle 3}$;

${\displaystyle \sqrt{x}-3=}$${\displaystyle 15:3}$;

${\displaystyle \sqrt{x}-3=}$${\displaystyle 5}$;

${\displaystyle \sqrt{x}=}$${\displaystyle 8}$;

${\displaystyle x=}$${\displaystyle 8^2}$;

${\displaystyle x=}$${\displaystyle 64}$;

ґ) ${\displaystyle \sqrt{14+5x}=}$${\displaystyle 8}$;

${\displaystyle 14+5x=}$${\displaystyle 8^2}$;

${\displaystyle 14+5x=}$${\displaystyle 64}$;

${\displaystyle 5x=}$${\displaystyle 64-14}$;

${\displaystyle 5x=}$${\displaystyle 50}$;

${\displaystyle x=}$${\displaystyle 50:5}$;

${\displaystyle x=}$${\displaystyle 10}$;

д) ${\displaystyle 1+\sqrt{1-x}=}$${\displaystyle 0}$;

${\displaystyle \sqrt{1-x}=}$${\displaystyle -1}$;

арифметичний корінь не може бути від’ємним — коренів немає.