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a fost răspuns

[tex] log_{5} \frac{4x^{2}-1 }{3-x}+ log_{5} \frac{3-x}{9+3x}=0[/tex]

Răspuns :

[tex] \frac{4 x^{2} -1}{3-x}* \frac{3-x}{9+3x}=1 \\ \frac{4 x^{2} -1}{9+3x}=1 \\ x \neq -3 \\ 4 x^{2} -1=9+3x \\ 4 x^{2} -3x-10=0 [/tex]
Δ=169
[tex] x_{1}= \frac{3-13}{8}=- \frac{5}{4} \\ x_{2}= \frac{3+13}{8}=2 [/tex]
S={2}
Teodoraalbei
㏒₅[(4x²-1)/(3-x)·(3-x)/(9+3x)]=0
se simplifica 3-x cu 3-x si obtinem
(4x²-1)/(9+3x)=1
4x²-1=9+3x
4x²-3x-10=0
x₁=2
x₂=-5/4

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