711 Вправа 711Алгебра 8 клас Тарасенкова 2025 · § 16 · стор. 207

711. Розв’яжіть графічно рівняння:

1) ${\displaystyle \sqrt{x}=}$${\displaystyle \frac{x}{3}}$; ${\displaystyle y=}$${\displaystyle \sqrt{x}}$ і ${\displaystyle y=}$${\displaystyle \frac{x}{3}}$.

${\displaystyle y=}$${\displaystyle \sqrt{x}}$:

${\displaystyle y=}$${\displaystyle \frac{x}{3}}$:

${\displaystyle x=}$${\displaystyle 0}$ і ${\displaystyle x=}$${\displaystyle 9}$;

2) ${\displaystyle \sqrt{x}=}$${\displaystyle x^2}$; ${\displaystyle y=}$${\displaystyle \sqrt{x}}$ і ${\displaystyle y=}$${\displaystyle x^2}$.

${\displaystyle y=}$${\displaystyle \sqrt{x}}$:

${\displaystyle y=}$${\displaystyle x^2}$:

${\displaystyle x=}$${\displaystyle 0}$ і ${\displaystyle x=}$${\displaystyle 1}$;

3) ${\displaystyle \sqrt{x}=}$${\displaystyle \frac{1}{x}}$; ${\displaystyle y=}$${\displaystyle \sqrt{x}}$ і ${\displaystyle y=}$${\displaystyle \frac{1}{x}}$.

${\displaystyle y=}$${\displaystyle \sqrt{x}}$:

${\displaystyle y=}$${\displaystyle \frac{1}{x}}$:

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

4) ${\displaystyle \sqrt{x}=}$${\displaystyle \frac{x}{2}}$; ${\displaystyle y=}$${\displaystyle \sqrt{x}}$ і ${\displaystyle y=}$${\displaystyle \frac{x}{2}}$.

${\displaystyle y=}$${\displaystyle \sqrt{x}}$:

${\displaystyle y=}$${\displaystyle \frac{x}{2}}$:

${\displaystyle x=}$${\displaystyle 0}$ і ${\displaystyle x=}$${\displaystyle 4}$.