Question

$\sqrt{x+1}$ + 1 = 4$x^2$ + $\sqrt{3x}$

Answer 1

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow4x^2-1+\sqrt{3x}-\sqrt{x+1}=0\)

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

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

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

\(\Leftrightarrow2x-1=0\) (do \(2x+1+\dfrac{1}{\sqrt{3x}+\sqrt{x+1}}>0;\forall x\ge0\))

\(\Leftrightarrow x=\dfrac{1}{2}\)