Matematyka
marek789
2017-06-23 16:14:39
Rozwiąż równanie 2x-4/x = x/2x-4 gdzie x =/= 0 i x =/= 2
Odpowiedź
samsung1702
2017-06-23 21:12:40

[latex]\ dfrac{2x-4}{x} = dfrac{x}{2x-4} /cdot x(2x-4) , xin Rackslash{0; 2} \ \(2x-4)^2=x^2 \ \(2x-4)^2-x^2=0 \ \(2x-4-x)(2x-4+x)=0 \ \(x-4)(3x-4)=0 \ \x-4=0 vee 3x-4=0 \ \xin{ frac{4}{3} ; 4}.[/latex]

xsandra
2017-06-23 21:13:55

[latex]frac{2x-4}{x}=frac{x}{2x-4)} |cdot x(2x-4)[/latex] [latex]x eq 0 i x eq 2\\D = R-lbrace0;2 brace[/latex] [latex](2x-4)^{2} = x^{2}\\4x^{2}-16x+16-x^{2}= 0\\3x^{2}-16x+16 =0\\Delta = (-16)^{2}-4cdot2cdot16 = 256-192 = 64\\sqrt{Delta} = 8\\x_1 = frac{-(-16)-8}{6} = frac{8}{6} = frac{4}{3}\\x_2 = frac{-(-16)+8}{6} = frac{24}{6} = 4\\xinlbracefrac{4}{3}:4 brace[/latex]

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