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)>0Do -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>0Vì vậy x-4>0\Rightarrow x>4Giao 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)>0Do
-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