Điều kiện $x +3 \ge 0$$pt(1)\Leftrightarrow (y^2+3)^2=\Bigg[(x+3)+\sqrt{x+3} \Bigg]^2\Leftrightarrow y^2=x+\sqrt{x+3}$Thế vào $pt(2)$, ta dc $(4x-1)\bigg(x+\sqrt{x+3}+\sqrt[3]{3x+5} \Bigg)=4x^2+3x+8$$\Leftrightarrow (4x-1) \bigg(\sqrt{x+3}+\sqrt[3]{x+5} \bigg)=4(x+2)$$\Leftrightarrow \sqrt{x+3}+\sqrt[3]{x+5}=\frac{4(x+2)}{4x-1} \quad (x \ne \frac 14) \quad (\bigstar)$Vì $\sqrt{x+3}+\sqrt[3]{x+5} >1\Leftrightarrow \frac{4(x+2)}{4x-1}>1\Leftrightarrow x> \frac 14$$(\bigstar)\Leftrightarrow \sqrt{x+3}-2+\sqrt[3]{x+5}-2=\frac{9}{4x-1}-3$$\Leftrightarrow \frac{x-1}{\sqrt{x+3}+2}+\frac{3(x-1)}{\sqrt[3]{(3x+5)^2}+2\sqrt[3]{3x+5}+4}+\frac{12(x-1)}{4x-1}=0$$\Leftrightarrow x=1$Nghiệm: $\begin{cases}x=1 \\ y=\mp\sqrt 3 \end{cases}$
Điều kiện $x +3 \ge 0$$pt(1)\Leftrightarrow (y^2+3)^2=\Bigg[(x+3)+\sqrt{x+3} \Bigg]^2\Leftrightarrow y^2=x+\sqrt{x+3}$Thế vào pt(2)...
Điều kiện $x +3 \ge 0$$pt(1)\Leftrightarrow (y^2+3)^2=\Bigg[(x+3)+\sqrt{x+3} \Bigg]^2\Leftrightarrow y^2=x+\sqrt{x+3}$Thế vào
$pt(2)
$, ta dc $(4x-1)\bigg(x+\sqrt{x+3}+\sqrt[3]{3x+5} \Bigg)=4x^2+3x+8$$\Leftrightarrow (4x-1) \bigg(\sqrt{x+3}+\sqrt[3]{x+5} \bigg)=4(x+2)$$\Leftrightarrow \sqrt{x+3}+\sqrt[3]{x+5}=\frac{4(x+2)}{4x-1} \quad (x \ne \frac 14) \quad (\bigstar)$Vì $\sqrt{x+3}+\sqrt[3]{x+5} >1\Leftrightarrow \frac{4(x+2)}{4x-1}>1\Leftrightarrow x> \frac 14$$(\bigstar)\Leftrightarrow \sqrt{x+3}-2+\sqrt[3]{x+5}-2=\frac{9}{4x-1}-3$$\Leftrightarrow \frac{x-1}{\sqrt{x+3}+2}+\frac{3(x-1)}{\sqrt[3]{(3x+5)^2}+2\sqrt[3]{3x+5}+4}+\frac{12(x-1)}{4x-1}=0$$\Leftrightarrow x=1$Nghiệm: $\begin{cases}x=1 \\ y=\mp\sqrt 3 \end{cases}$