647° Вправа 647Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 192

647. Розв’яжіть рівняння:

1) ${\displaystyle \sqrt{x}=}$${\displaystyle 2}$; ${\displaystyle x=}$${\displaystyle 2^2}$; ${\displaystyle x=}$${\displaystyle 4}$;

2) ${\displaystyle \sqrt{x}=}$${\displaystyle 10}$; ${\displaystyle x=}$${\displaystyle 100}$;

3) ${\displaystyle 2\sqrt{x}-6=}$${\displaystyle 0}$; ${\displaystyle \sqrt{x}=}$${\displaystyle 3}$; ${\displaystyle x=}$${\displaystyle 9}$;

4) ${\displaystyle \sqrt{x-2}-9=}$${\displaystyle 0}$; ${\displaystyle \sqrt{x-2}=}$${\displaystyle 9}$; ${\displaystyle x-2=}$${\displaystyle 81}$; ${\displaystyle x=}$${\displaystyle 83}$;

5) ${\displaystyle \sqrt{x-12}=}$${\displaystyle 7}$; ${\displaystyle x-12=}$${\displaystyle 49}$; ${\displaystyle x=}$${\displaystyle 61}$;

6) ${\displaystyle \sqrt{x+21}=}$${\displaystyle -3<0}$; коренів немає;

7) ${\displaystyle \sqrt{x+3}+4=}$${\displaystyle 0}$; ${\displaystyle \sqrt{x+3}=}$${\displaystyle -4<0}$; коренів немає;

8) ${\displaystyle 3\sqrt{x+9}=}$${\displaystyle 0}$; ${\displaystyle x+9=}$${\displaystyle 0}$; ${\displaystyle x=}$${\displaystyle -9}$;

9) ${\displaystyle -2\sqrt{x+9}+6=}$${\displaystyle 0}$; ${\displaystyle \sqrt{x+9}=}$${\displaystyle 3}$; ${\displaystyle x+9=}$${\displaystyle 9}$; ${\displaystyle x=}$${\displaystyle 0}$.