671 Вправа 671Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 196

671. Розв’яжіть рівняння:

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

${\displaystyle {\left(\sqrt{x}\right)}^2-1^2=}$${\displaystyle 2-x}$;

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

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

${\displaystyle 2x=}$${\displaystyle 3}$ | ${\displaystyle :2}$

${\displaystyle x=}$${\displaystyle 1,5}$;

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

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

${\displaystyle x+x=}$${\displaystyle 5+3}$;

${\displaystyle 2x=}$${\displaystyle 8}$ | ${\displaystyle :2}$

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

3) ${\displaystyle (\sqrt{x}+2)(\sqrt{x}-2)=}$${\displaystyle (\sqrt{x}-4)(\sqrt{x}+5)}$;

${\displaystyle {\left(\sqrt{x}\right)}^2-2^2=}$${\displaystyle x+5\sqrt{x}-4\sqrt{x}-20}$;

${\displaystyle x-4=}$${\displaystyle x+\sqrt{x}-20}$;

${\displaystyle x-x-\sqrt{x}=}$${\displaystyle -20+4}$;

${\displaystyle -\sqrt{x}=}$${\displaystyle -16}$ | ${\displaystyle :\left(-1\right)}$

${\displaystyle \sqrt{x}=}$${\displaystyle 16}$;

${\displaystyle x=}$${\displaystyle {16}^2}$;

${\displaystyle x=}$${\displaystyle 256}$;

4) ${\displaystyle (2\sqrt{x}-7)(2\sqrt{x}+7)=}$${\displaystyle (4\sqrt{x}-6)(\sqrt{x}+5)}$;

${\displaystyle {\left(2\sqrt{x}\right)}^2-7^2=}$${\displaystyle 4x+20\sqrt{x}-6\sqrt{x}-30}$;

${\displaystyle 4x-49=}$${\displaystyle 4x+14\sqrt{x}-30}$;

${\displaystyle 4x-4x-14\sqrt{x}=}$${\displaystyle -30+49}$;

${\displaystyle -14\sqrt{x}=}$${\displaystyle 19}$ | ${\displaystyle :\left(-14\right)}$

${\displaystyle \sqrt{x}=}$${\displaystyle -\frac{19}{14}}$;

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

5) ${\displaystyle {(\sqrt{x}-5)}^2=}$${\displaystyle 26+x}$;

${\displaystyle {\left(\sqrt{x}\right)}^2-2\cdot \sqrt{x}\cdot 5+5^2=}$${\displaystyle 26+x}$;

${\displaystyle x-10\sqrt{x}+25=}$${\displaystyle 26+x}$;

${\displaystyle x-x-10\sqrt{x}=}$${\displaystyle 26-25}$;

${\displaystyle -10\sqrt{x}=}$${\displaystyle 1}$ | ${\displaystyle :\left(-10\right)}$

${\displaystyle \sqrt{x}=}$${\displaystyle -\frac{1}{10}}$;

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

6) ${\displaystyle {(\sqrt{x}-2\sqrt{2})}^2=}$${\displaystyle 8+x}$;

${\displaystyle {\left(\sqrt{x}\right)}^2-2\cdot \sqrt{x}\cdot 2\sqrt{2}+{\left(2\sqrt{2}\right)}^2=}$${\displaystyle 8+x}$;

${\displaystyle x-4\sqrt{2x}+8=}$${\displaystyle 8+x}$;

${\displaystyle x-x-4\sqrt{2x}=}$${\displaystyle 8-8}$;

${\displaystyle -4\sqrt{2x}=}$${\displaystyle 0}$ | ${\displaystyle :\left(-4\right)}$

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

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

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