672 Вправа 672Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 196

672. Розв’яжіть рівняння:

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

${\displaystyle {\left(\sqrt{x}\right)}^2-{\left(\sqrt{2}\right)}^2=}$${\displaystyle 2-x}$;

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

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

${\displaystyle 2x=}$${\displaystyle 4}$ | ${\displaystyle :2}$

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

2) ${\displaystyle (\sqrt{x}+5)(\sqrt{x}-5)=}$${\displaystyle (\sqrt{x}-7)(\sqrt{x}+6)}$;

${\displaystyle {\left(\sqrt{x}\right)}^2-5^2=}$${\displaystyle x+6\sqrt{x}-7\sqrt{x}-42}$;

${\displaystyle x-25=}$${\displaystyle x-\sqrt{x}-42}$;

${\displaystyle x-x+\sqrt{x}=}$${\displaystyle -42+25}$;

${\displaystyle \sqrt{x}=}$${\displaystyle -17}$;

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

3) ${\displaystyle {(\sqrt{x}-\sqrt{15})}^2-x=}$${\displaystyle -30}$;

${\displaystyle {\left(\sqrt{x}\right)}^2-2\cdot \sqrt{x}\cdot \sqrt{15}+{\left(\sqrt{15}\right)}^2-x=}$${\displaystyle -30}$;

${\displaystyle x-2\sqrt{15x}+15-x=}$${\displaystyle -30}$;

${\displaystyle -2\sqrt{15x}=}$${\displaystyle -30-15}$;

${\displaystyle -2\sqrt{15x}=}$${\displaystyle -45}$ | ${\displaystyle :\left(-2\right)}$

${\displaystyle \sqrt{15x}=}$${\displaystyle 22,5}$;

${\displaystyle 15x=}$${\displaystyle {22,5}^2}$;

${\displaystyle 15x=}$${\displaystyle 506,25}$ | ${\displaystyle :15}$

${\displaystyle x=}$${\displaystyle 33,75}$.