Question

Giaỉ phương trình sau: \(\sqrt{4x+1}-\sqrt{3-x}+4x^2-5x-8=0\)

Answer 1

ĐKXĐ: \(-\dfrac{1}{4}\le x\le3\)

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

\(\Leftrightarrow\dfrac{4\left(x-2\right)}{\sqrt{4x+1}+3}+\dfrac{x-2}{1+\sqrt{3-x}}+\left(x-2\right)\left(4x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{1}{1+\sqrt{3-x}}+4x+3\right)=0\)

Do \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{1}{1+\sqrt{3-x}}+4x+3>0;\forall x\in\left[-\dfrac{1}{4};3\right]\)

\(\Rightarrow x-2=0\Rightarrow x=2\)