(GV) Funções
Enviado: Sex 25 Jan, 2013 10:41
Resolvendo a desigualdade [tex3]1-3x > \sqrt{2 + x^{2} -3}[/tex3]
a)[tex3]x<\frac{3-\sqrt{41}}{16}[/tex3]
b)[tex3]x\leq \frac{1}{3}[/tex3]
c)[tex3]x<1 \,\,\cup \,\,x>2[/tex3]
d)[tex3]\frac{1}{3}\leq x\leq \frac{3+\sqrt{41}}{16}[/tex3]
e)[tex3]x<\frac{3-\sqrt{41}}{16}\,\,\cup\,\,x>\frac{3+\sqrt{41}}{16}[/tex3]
obtemos:a)[tex3]x<\frac{3-\sqrt{41}}{16}[/tex3]
b)[tex3]x\leq \frac{1}{3}[/tex3]
c)[tex3]x<1 \,\,\cup \,\,x>2[/tex3]
d)[tex3]\frac{1}{3}\leq x\leq \frac{3+\sqrt{41}}{16}[/tex3]
e)[tex3]x<\frac{3-\sqrt{41}}{16}\,\,\cup\,\,x>\frac{3+\sqrt{41}}{16}[/tex3]