Намерете корените на уравнението...

[Гост]

( \frac {x-3}{ \sqrt{3} } ) [2] = ( \frac{x-1}{ -\sqrt{2} } [2]

[Гост]

[tex](\frac {x-3}{\sqrt{3}})[2] = (\frac{x-1}{-\sqrt{2}}[2][/tex]

[Гост]

[tex](\frac {x-3}{\sqrt{3}})[2] - степен = (\frac{x-1}{-\sqrt{2}})[2] - степен[/tex]

[Гост]

[tex](\frac {x-3}{\sqrt{3}})[2] = (\frac{x-1}{-\sqrt{2}})[2][/tex] това [2] е степен

[ganka simeonova]

[tex]( \frac {x-3}{ \sqrt{3} } )^2 = ( \frac{x-1}{ -\sqrt{2} })^2[/tex]

[Гост]

Ако може решение?

[Xixibg]

[tex]( \frac {x-3}{ \sqrt{3} } )^2 = ( \frac{x-1}{ -\sqrt{2} })^2[/tex]
[tex]\frac{x^2-6x+9}{3}=\frac{x^2-2x+1}{2}[/tex]
[tex]2x^2-12x+18=3x^2-6x+3[/tex]
[tex]x^2+6x-15=0[/tex]
[tex](x-(-3+2\sqrt{6}))(x-(-3-2\sqrt{6}))=0[/tex]
[tex]x_1=-3+2\sqrt{6} ; x_2=-3-2\sqrt{6}[/tex]