Question

f(x)=3x-1/9x^2-1

решите пожалуйста

Answer 1

х=$\frac{3x-1}{9x^{2} -1 }$ =  x = $\frac{3x-1}{9x^{2} -1}$, х≠$\frac{1}{3}$, х≠$-\frac{1}{3}$

х =  $\frac{3x-1}{(3x-1)* (3x+1)}$

x = $\frac{1}{3x+1}$

(3x+1) x=1

(3x+1) x-1 =  0

3x²+x=-1=0

x= $\frac{-1+-\sqrt{1^{2} - 4*3*(-1)} }{2*3}$

x = $\frac{-1+-\sqrt{1+12} }{6}$

x = $\frac{-1+-\sqrt{13} }{6}$

$x=\frac{-1+\sqrt{13} }{6}$

$x = \frac{-1-\sqrt{13} }{6}$, х≠$\frac{1}{3}$, х≠-$\frac{1}{3}$

х = $\frac{-1+\sqrt{13} }{6}$

х = $\frac{-1-\sqrt{13} }{6}$

х₁ = $\frac{-1-\sqrt{13} }{6}$, х₂ = $\frac{-1+\sqrt{13} }{6}$