ĐK \(\left\{\begin{matrix} x^2+4x+3\geq 0\\ x+1\geq 0\\ 4x^2-3\geq 0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -1\\ \bigg \lbrack\begin{matrix} x\leq -\frac{\sqrt{3}}{2}\\ x\geq \frac{\sqrt{3}}{2} \end{matrix}\\ \end{matrix}\right.\Leftrightarrow \bigg \lbrack\begin{matrix} -1\leq x\leq -\frac{\sqrt{3}}{2}\\ x\geq \frac{\sqrt{3}}{2} \ \ \ \ \ \ \ \ \ \end{matrix}\)
Ta có (pt1)
\(\Leftrightarrow \sqrt{x+1}.\sqrt{x+3}+y-y\sqrt{x+3}=y^3+\sqrt{x+1}-y^2\sqrt{x+1}\)
\(\Leftrightarrow \sqrt{x+3}(\sqrt{x+1}-y)+(y-\sqrt{x+1})=y^2(y-\sqrt{x+1})\)
\(\Leftrightarrow (y-\sqrt{x+1})(y^2+\sqrt{x+3}-1)=0\Leftrightarrow y-\sqrt{x+1}=0\)
(do \(y^2+\sqrt{x+1}(y^2+\sqrt{x+3}-1);\forall x\geq -1\))
\(y-\sqrt{x+1}=0\Leftrightarrow y=\sqrt{x+1}\Leftrightarrow \left\{\begin{matrix} x\geq 0 \\ y^2=x+1 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} y\geq 0\\ y^2-1=x \end{matrix}\right.\)
Thế vào (pt2) ta được \(2x^2(3x^2+1)=(x^2+1)(1-3x\sqrt{4x^2-3})(pt*)\)
TH1: Với \(x\geq \frac{\sqrt{3}}{2}\)
PT(*) \(\Leftrightarrow \frac{6x^4+2x^2}{x^2+1}=1-3x\sqrt{4x^2-3}\) (Phương trình vô nghiệm)
Do \(VT=\frac{6x^4+2x^2}{x^2+1}\geq \frac{\frac{27}{8}+2x^2}{x^2+1}> \frac{2+2x^2}{x^2+1}=2\)
còn \(VP=1-3x\sqrt{4x^2-3}\leq 1\)
TH2: Với \(-1\leq x\leq -\frac{\sqrt{3}}{2}\)
PT(*) \(\Leftrightarrow 6x^4+2x^2=(x^2+1)(1+3\sqrt{4x^4-3x^2})\) Chia hai vế pt này cho x4 ta được:
\(\Leftrightarrow 6+\frac{2}{x^2}=\left ( 1+\frac{1}{x^2} \right )\left ( \frac{1}{x^2+3\sqrt{4-\frac{3}{x^2}}} \right )\)
Đặt \(t=\frac{1}{x^2};\left ( t\in \left [ 1;\frac{4}{3} \right ] \right )\)
Ta được pt: \(6+2t=(1+t)(t+3\sqrt{4-3t})\Leftrightarrow \frac{2t+6}{t+1}=t+3\sqrt{4-3t}\)
\(\Leftrightarrow \frac{2t+6}{t+1}-4=t-1+3(\sqrt{4-3t}-1)\Leftrightarrow \frac{2-2t}{t+1}=t-1+\frac{3(3-3t)}{\sqrt{4-3t}+1}\)
\(\Leftrightarrow (1-t)\left ( \frac{2}{t+1}+1 \right )=\frac{9(1-t)}{\sqrt{4-3t}+1}\Leftrightarrow \bigg \lbrack\begin{matrix} 1-t=0\\ \frac{2}{t+1}+1=\frac{9}{\sqrt{4-3t}+1} \end{matrix}\)
*) Pt \(1-t=0\Leftrightarrow t=1\) được \(x=-1\Rightarrow y=0\)
*) Pt \(\frac{2}{t+1}+1=\frac{9}{\sqrt{4-3t}+1}\) vô nghiệm vì \(t\geq 1\Rightarrow \left\{\begin{matrix} \frac{2}{t+1}+1\leq 2\\\\ \frac{9}{\sqrt{4-3t}+1}\geq \frac{9}{2} \end{matrix}\right.\)

Hệ có đúng một nghiệm: (x; y) = (-1; 0)