Điều kiện x+3≥0pt(1)⇔(y2+3)2=[(x+3)+√x+3]2⇔y2=x+√x+3Thế vào pt(2)...
Điều kiện
x+3≥0pt(1)⇔(y2+3)2=[(x+3)+√x+3]2⇔y2=x+√x+3Thế vào
$pt(2)
,tadc(4x-1)\bigg(x+\sqrt{x+3}+\sqrt[3]{3x+5} \Bigg)=4x^2+3x+8⇔(4x−1)(√x+3+3√x+5)=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(★)⇔√x+3−2+3√x+5−2=94x−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: $\begin{cases}x=1 \\ y=\mp\sqrt 3 \end{cases}$