Odpowiedź:
a.
[tex]f(x) =\frac{x}{\sqrt{2-x } } \\0 = \frac{x}{\sqrt{2-x} } \\0 - \frac{x}{\sqrt{2-x} } \\-\frac{x}{\sqrt{2-x} } =0\\\frac{x}{\sqrt{2-x} }=0\\x=0\\0=\frac{x}{\sqrt{2-0} } \\0=0\\x=0\\[/tex]
b.
[tex]f(x)=\frac{\sqrt{x}+40 }{x^{2} -25} \\ 0=\frac{\sqrt{x} +10}{x^{2} -25}\\0-\frac{\sqrt{x} +10}{x^{2} -25}=0\\\\-\frac{\sqrt{x} +10}{x^{2} -25} =0\\\frac{\sqrt{x} +10}{x^{2} -25}=0\\\sqrt{x} +10=0\\\sqrt{x} =-10\\[/tex]
=×∈∅
