Răspuns:
Mai intai DVA
x+4≠0
x≠-4
(x²-5x)/(x+4)>0
x²-5x≠0
x≠{0,5}
tabel
x l -∞............-4.......0........5................+∞
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x²-5xl + + + 0- - 0+ + +
_________________________________________________________x+4 l - - -0+ + + + + +
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(x²-5x)/(x+4)l- - -l+ ++l- - -l+ + +
consideri intervalele pozitive
x∈(-4,0)U(5,+∞)
Pt ca baza primului logaritm este subunitara (1/5) pui conditia ca logaritmul in baza 7 sa fie supra unitar.Adica
(x²-5x)/(x+4)>7
(x²-5x)/(x+4)-7>0
(x²-5x-7x-28)/(x+4)>0
(x²-12x-28)/(X+4)>0
x²-12x-28=0
x1= -2
x2=14
Tabelul de semne
x l-∞....................-4........-2............>..................+14..................+∞
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x²-12x-28l+++++++ l-------------------------------------0+++++++++++
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(x²-12x-28)/(x+4)l+++l-------------------------------------0+++++++++++++++
Cosideri valorile pozitive strict
x∈(-∞, -4)U(14.+∞)
Intersectezi acest interval cu DVA si aflii solutiile
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x+4 l----------------------0+++++++++++++++++++++++++++++++++++++
x²
Explicație pas cu pas:
