Điều kiện : $-\frac{4}{3}\leq x\leq 5$$(\sqrt{3x+4}-4)+(1-\sqrt{5-x})+3x^2-8x-19+3>0$$\frac{3x+4-16}{\sqrt{3x+4}+4}+\frac{1-5+x}{1+\sqrt{5-x}}+(x-4)(3x+4)>0$$(x-4)(\frac{3}{\sqrt3x+4}+\frac{1}{1+\sqrt{5-x}}+3x+4)>0$Do $-4\leq 3x \leq 15$ nên $0\leq 3x+4\leq 19$ Suy ra: $\frac{3}{\sqrt{3x+4}+4}+\frac{1}{1+\sqrt{5-x}}+3x+4>0$Vì vậy $x-4>0\Rightarrow x>4$Giao điều kiện: $4
Điều kiện : $-\frac{4}{3}\leq x\leq 5$$(\sqrt{3x+4}-4)+(1-\sqrt{5-x})+3x^2-8x-19+3>0$$\frac{3x+4-16}{\sqrt{3x+4}+4}+\frac{1-5+x}{1+\sqrt{5-x}}+(x-4)(3x+4)>0$$(x-4)(\frac{3}{\sqrt3x+4}+\frac{1}{1+\sqrt{5-x}}+3x+4)>0$Do $-4\leq 3x \leq 15$ nên $0\leq 3x+4\leq 19$ Suy ra: $\frac{3}{\sqrt{3x+4}+4}+\frac{1}{\sqrt{5-x}}+3x+4>0$Vì vậy $x-4>0\Rightarrow x>4$Giao điều kiện: $4<x\leq 5$
Điều kiện : $-\frac{4}{3}\leq x\leq 5$$(\sqrt{3x+4}-4)+(1-\sqrt{5-x})+3x^2-8x-19+3>0$$\frac{3x+4-16}{\sqrt{3x+4}+4}+\frac{1-5+x}{1+\sqrt{5-x}}+(x-4)(3x+4)>0$$(x-4)(\frac{3}{\sqrt3x+4}+\frac{1}{1+\sqrt{5-x}}+3x+4)>0$Do $-4\leq 3x \leq 15$ nên $0\leq 3x+4\leq 19$ Suy ra: $\frac{3}{\sqrt{3x+4}+4}+\frac{1}{
1+\sqrt{5-x}}+3x+4>0$Vì vậy $x-4>0\Rightarrow x>4$Giao điều kiện: $4