669 Вправа 669Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 195

669. Розв’яжіть рівняння:

1) ${\displaystyle \sqrt{x}+2\sqrt{x}=}$${\displaystyle 9}$;

${\displaystyle (1+2)\sqrt{x}=}$${\displaystyle 9}$;

${\displaystyle 3\sqrt{x}=}$${\displaystyle 9}$ | ${\displaystyle :3}$

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

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

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

2) ${\displaystyle 9\sqrt{x}-5\sqrt{x}=}$${\displaystyle 4}$;

${\displaystyle (9-5)\sqrt{x}=}$${\displaystyle 4}$;

${\displaystyle 4\sqrt{x}=}$${\displaystyle 4}$ | ${\displaystyle :4}$

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

${\displaystyle x=}$${\displaystyle 1^2}$;

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

3) ${\displaystyle \sqrt{4x}-\sqrt{9x}=}$${\displaystyle 9}$;

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

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

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

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

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

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

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

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

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

5) ${\displaystyle \sqrt{x+3}-5\sqrt{x+3}+4=}$${\displaystyle 0}$;

${\displaystyle (1-5)\sqrt{x+3}=}$${\displaystyle -4}$;

${\displaystyle -4\sqrt{x+3}=}$${\displaystyle -4}$ | ${\displaystyle :\left(-4\right)}$

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

${\displaystyle x+3=}$${\displaystyle 1^2}$;

${\displaystyle x=}$${\displaystyle 1-3}$;

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

6) ${\displaystyle \sqrt{x+1}+3\sqrt{x+1}=}$${\displaystyle 8}$;

${\displaystyle (1+3)\sqrt{x+1}=}$${\displaystyle 8}$;

${\displaystyle 4\sqrt{x+1}=}$${\displaystyle 8}$ | ${\displaystyle :4}$

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

${\displaystyle x+1=}$${\displaystyle 2^2}$;

${\displaystyle x=}$${\displaystyle 4-1}$;

${\displaystyle x=}$${\displaystyle 3}$.