Điều kiện x+3≥0
pt(1)⇔(y2+3)2=[(x+3)+√x+3]2⇔y2=x+√x+3Thế vào pt(2), ta dc
(4x−1)(x+√x+3+3√3x+5)=4x2+3x+8
⇔(4x−1)(√x+3+3√x+5)=4(x+2)
⇔√x+3+3√x+5=4(x+2)4x−1(x≠14)(★)
Vì √x+3+3√x+5>1⇔4(x+2)4x−1>1⇔x>14
(★)⇔√x+3−2+3√x+5−2=94x−1−3
⇔x−1√x+3+2+3(x−1)3√(3x+5)2+23√3x+5+4+12(x−1)4x−1=0
⇔x=1
Nghiệm: {x=1y=∓√3