Question

Помогите пожалуйста!

Answer 1

$y=\frac{\sqrt{4x+1} }{x-3}\\\\1)A(2 \ ; \ -3) \ \Rightarrow \ x=2 \ ; \ y= - 3\\\\-3=\frac{\sqrt{4\cdot2+1} }{2-3}\\\\-3=\frac{\sqrt{9} }{-1}\\\\-3=-3-verno\\\\\\2)B(0 \ ; \ -3) \ \Rightarrow \ x=0 \ ; \ y= - 3\\\\-3=\frac{\sqrt{4\cdot0+1} }{0-3} \\\\-3=\frac{1}{-3}\\\\-3=-\frac{1}{3}-neverno\\\\\\3)C(6 \ ; \ 2) \ \Rightarrow \ x=6 \ ; \ y= 2$

$2=\frac{\sqrt{4\cdot6+1} }{6-3} \\\\2=\frac{\sqrt{25} }{3}\\\\2=\frac{5}{3}-neverno\\\\\\4)D(20 \ ; \ 17) \ \Rightarrow \ x=20 \ ; \ y= 17\\\\17=\frac{\sqrt{4\cdot20+1} }{20-3}\\\\17=\frac{\sqrt{81} }{17}\\\\17=\frac{9}{17} -neverno\\\\Otvet:\boxed{A(2 \ ; \ -3)}$