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Aratati ca E(x)=[tex] \frac{ x^{2}-9 }{ x^{2}-4x+3 } [/tex]+[tex] \frac{x+2}{x+1} - \frac{ x^{2}+5x+2 }{ x^{2}-1} =1[/tex] , pt orice x∈R\{-1;1;3}

Răspuns :

Miky93id
vine (x-3)(x+3) supra x la a2a -x-3x+3 +x+2 supra x+1  - x la a2a +5x +2 supra (x-1)(x+1)
(x-3)(x+3) supra x(x-1)-3(x-1)  +x+2 supra x+1 - x la a2a +5x+2 supra (x-1)(x+1)
(x-3)(x+3)supra(x-1)(x-3) + tot sirul... simplificam in prima ecuatie (x-3) si obtinem
(x+3) supra (x-1) + x+2 supra x+1  - x la a2a +5x+2 supra (x-1)(x+1) aducem la acelasi numitor si obtinem
(x+3)(x+1)+(x+2)(x-1)-x la a2a -5x -2
x la a2a+x+3x+3+x la a2a -x+2x-2 - x la a2a -5x-2
x la a2a-1=(x-1)(x+1)=1 rezulta ca x-1=1 x=2    sau  x+1=1  x=0

Danaradu70id
x^2 - 4x+3=x^2-3x-x+3=x(x-1)-3(x-1)=(x-3)(x-1)
[tex] x^{2} [/tex] - 1 =(x-1)(x+1)
[tex] x^{2} [/tex] - 9=(x-3)(x+3)
E(x)=[tex] \frac{(x-3)(x+3)}{(x-3)(x-1)} [/tex] + [tex] \frac{x+2}{x+1} [/tex] - [tex] \frac{x^2+5x+2}{x^2-1} [/tex] = 1
aducem la acelasi numitor
numitorul comun este (x-1)(x+1)=[tex] x^{2} [/tex] - 1
[tex] \frac{(x+1)(x+3)+(x+2)(x-1)-( x^{2} +5x+2)}{ x^{2} -1} [/tex] = [tex] \frac{ x^{2} +3x+x+3+x^{2} -x+2x-2- x^{2} -5x-2}{ x^{2} -1} [/tex]=[tex] \frac{ x^{2} -1}{ x^{2} -1} [/tex]=1

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