666 Вправа 666Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 195

666. Розв’яжіть рівняння:

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

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

${\displaystyle x=}$${\displaystyle 0}$ або ${\displaystyle x=}$${\displaystyle -4}$;

2) ${\displaystyle {(x-9)}^2=}$${\displaystyle 100}$;

${\displaystyle x-9=}$${\displaystyle 10}$ або ${\displaystyle x-9=}$${\displaystyle -10}$;

${\displaystyle x=}$${\displaystyle 19}$ або ${\displaystyle x=}$${\displaystyle -1}$;

3) ${\displaystyle {(x-25)}^2=}$${\displaystyle 0}$;

${\displaystyle x-25=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 25}$;

4) ${\displaystyle {(x+4)}^2-3=}$${\displaystyle 0}$;

${\displaystyle {(x+4)}^2=}$${\displaystyle 3}$;

${\displaystyle x+4=}$${\displaystyle \sqrt{3}}$ або ${\displaystyle x+4=}$${\displaystyle -\sqrt{3}}$;

${\displaystyle x=}$${\displaystyle -4+\sqrt{3}}$ або ${\displaystyle x=}$${\displaystyle -4-\sqrt{3}}$;

5) ${\displaystyle {(x-7)}^2-11=}$${\displaystyle 0}$;

${\displaystyle {(x-7)}^2=}$${\displaystyle 11}$;

${\displaystyle x-7=}$${\displaystyle \sqrt{11}}$ або ${\displaystyle x-7=}$${\displaystyle -\sqrt{11}}$;

${\displaystyle x=}$${\displaystyle 7+\sqrt{11}}$ або ${\displaystyle x=}$${\displaystyle 7-\sqrt{11}}$;

6) ${\displaystyle {(x+7)}^2+49=}$${\displaystyle 0}$;

${\displaystyle {(x+7)}^2=}$${\displaystyle -49<0}$;

коренів немає.