ĐKXĐ :
$\left\{ \begin{matrix}4x+1 \geq 0\\3x+4 \geq 0\end{matrix} \right. ⇔\left\{ \begin{matrix}x \geq \dfrac{-1}{4}\\x \geq \dfrac{-4}{3}\end{matrix} \right. ⇒ x \geq -\dfrac{1}{4}$
$\sqrt{4x+1}-\sqrt{3x+4}=1$
$⇔\sqrt{4x+1}=\sqrt{3x+4}+1$
$⇔4x+1=3x+4+2\sqrt{3x+4}+1$
$⇔4x-3x=4+2\sqrt{3x+4}+1-1$
$⇔x-4=2\sqrt{3x+4}$
$⇔(x-4)^2=(2\sqrt{3x+4})^2$
$⇔x^2-8x+16=4(3x+4)$
$⇔x^2-8x+16=12x+16$
$⇔x^2-20x=0$
$⇔x(x-20)=0$
$⇔\left[ \begin{array}{1}x=0\\x=20\end{array} \right.$
Thử lại :
$+) x=0 :$
$\sqrt{4.0+1}-\sqrt{3.0+4}=\sqrt{1}-\sqrt{4}=1-2=-1$ ( Loại )
$+) x=20 :$
$\sqrt{4.20+1}-\sqrt{3.20+4}=\sqrt{81}-\sqrt{64}=9-8=1 ( t/m )$
Vậy $S=\{ 20 \}$
