Đáp án:
$S=∅$
Giải thích các bước giải:
ĐKXĐ: $x\ge \dfrac{1}{2}$
$\sqrt{3x+4}-\sqrt{2x-1}=\sqrt{x+3}$
$⇔\sqrt{3x+4}=\sqrt{2x-1}+\sqrt{x+3}$
$⇔3x+4=2x-1+x+3+2\sqrt{(2x-1)(x+3)}$
$⇔2=2\sqrt{(2x-1)(x+3)}$
$⇔\sqrt{2x^2+5x-3}=1$
$⇔2x^2+5x-3=1$
$⇔2x^2+5x-4=0$
$Δ=5^2-4.2.(-4)=57$
$⇒\left[ \begin{array}{l}x=\dfrac{-5+\sqrt{57}}{4}\,(L)\\x=\dfrac{-5-\sqrt{57}}{4}\,(L)\end{array} \right.$
Vậy $S=∅$.
