670 Вправа 670Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 195

670. Розв’яжіть рівняння:

1) ${\displaystyle 3\sqrt{x}+2\sqrt{x}=}$${\displaystyle 10}$;

${\displaystyle (3+2)\sqrt{x}=}$${\displaystyle 10}$;

${\displaystyle 5\sqrt{x}=}$${\displaystyle 10}$ | ${\displaystyle :5}$

${\displaystyle \sqrt{x}=}$${\displaystyle 2}$;

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

${\displaystyle x=}$${\displaystyle 4}$;

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

${\displaystyle (1-3)\sqrt{x}=}$${\displaystyle 6}$;

${\displaystyle -2\sqrt{x}=}$${\displaystyle 6}$ | ${\displaystyle :\left(-2\right)}$

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

коренів немає, бо ${\displaystyle \sqrt{x}\ge 0}$;

3) ${\displaystyle \sqrt{16x}-\sqrt{25x}=}$${\displaystyle 0}$;

${\displaystyle 4\sqrt{x}-5\sqrt{x}=}$${\displaystyle 0}$;

${\displaystyle (4-5)\sqrt{x}=}$${\displaystyle 0}$;

${\displaystyle -\sqrt{x}=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 0^2}$;

${\displaystyle x=}$${\displaystyle 0}$;

4) ${\displaystyle \sqrt{x+5}-7\sqrt{x+5}+12=}$${\displaystyle 0}$;

${\displaystyle (1-7)\sqrt{x+5}=}$${\displaystyle -12}$;

${\displaystyle -6\sqrt{x+5}=}$${\displaystyle -12}$ | ${\displaystyle :\left(-6\right)}$

${\displaystyle \sqrt{x+5}=}$${\displaystyle 2}$;

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

${\displaystyle x=}$${\displaystyle 4-5}$;

${\displaystyle x=}$${\displaystyle -1}$.