704 Вправа 704Алгебра 8 клас Тарасенкова 2025 · § 16 · стор. 206

704. Знайдіть такі точки графіка функції y = √x, у яких:

1) ${\displaystyle x=}$${\displaystyle y}$, тобто ${\displaystyle \sqrt{x}=}$${\displaystyle x}$;

${\displaystyle \sqrt{x}-{\left(\sqrt{x}\right)}^2=}$${\displaystyle 0}$;

${\displaystyle \sqrt{x}(1-\sqrt{x})=}$${\displaystyle 0}$;

${\displaystyle \sqrt{x}=}$${\displaystyle 0}$ або ${\displaystyle \sqrt{x}=}$${\displaystyle 1}$;

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

точки ${\displaystyle \left(0;0\right)}$ і ${\displaystyle \left(1;1\right)}$;

2) ${\displaystyle y=}$${\displaystyle \sqrt{x}}$ і ${\displaystyle 2x+3y=}$${\displaystyle 27}$, тобто ${\displaystyle 2x+3\sqrt{x}-27=}$${\displaystyle 0}$;

${\displaystyle 2{\left(\sqrt{x}\right)}^2-6\sqrt{x}+9\sqrt{x}-27=}$${\displaystyle 0}$;

${\displaystyle 2\sqrt{x}(\sqrt{x}-3)+9(\sqrt{x}-3)=}$${\displaystyle 0}$;

${\displaystyle (\sqrt{x}-3)(2\sqrt{x}+9)=}$${\displaystyle 0}$;

${\displaystyle 2\sqrt{x}+9>0}$, тому ${\displaystyle \sqrt{x}-3=}$${\displaystyle 0}$;

${\displaystyle \sqrt{x}=}$${\displaystyle 3}$; ${\displaystyle x=}$${\displaystyle 9}$; ${\displaystyle y=}$${\displaystyle 3}$;

точка ${\displaystyle \left(9;3\right)}$.