Question

Answer 1

ĐKXĐ: \(-\dfrac{1}{3}\le x\le6\)

\(\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14x-5\right)< 0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)< 0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)< 0\)

Do \(\dfrac{3}{\sqrt{3x+1}+1}+\dfrac{1}{1+\sqrt{6-x}}+3x+1>0;\forall x\in\left[-\dfrac{1}{3};6\right]\) nên BPT tương đương:

\(x-5< 0\Leftrightarrow x< 5\)

Vậy tập nghiệm của BPT là: \(-\dfrac{1}{3}\le x< 5\)