Đáp án:
$ĐK: x ≥-1 $
$\sqrt[]{2x+3}+\sqrt[]{x+1}=3x +2\sqrt[]{(2x+3).(x+1)}-16$ (1)
Đặt $\sqrt[]{2x+3}=a$ và $\sqrt[]{x+1}=b$ ($a,b\geq 0$)
⇒a²=2x+3; b²=x+1
⇒a²+b²=3x+4
⇒a²+b²-20=3x-16
Khi đó: (1) ⇔ a+b=a²+b²-20+2ab
⇔a+b=(a+b)²-20
⇔(a+b)²-(a+b)-20=0
⇔(a+b+4).(a+b-5)=0
⇔\(\left[ \begin{array}{l}a+b+4=0\\a+b-5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}a+b=-4\\a+b=5\end{array} \right.\)
Vì a,b≥0 nên a+b≥0 ⇒a+b=5⇒b=5-a
Vì a²=2x+3; b²=x+1 nên a²-2b²=1⇒a²-2(5-a)²=1⇒a²-2(25-10a+a²)=1
⇒a²-50+20a-2a²-1=0
⇒-a²+20a-51=0
⇒\(\left[ \begin{array}{l}a=3(TM)\\a=17(TM)\end{array} \right.\)
TH1: a=3⇒b=5-a=5-3=2(TM)
⇒$\sqrt[]{2x+3}=3$⇒$2x+3=9$⇒$2x=6$⇒$x=3 (TM)$
TH2:a=17⇒b=5-a=5-17=-12(KTM)
Vậy x=3
