ĐKXĐ : \(x\ge-3\)
+ Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+3}\ge0\\b=\sqrt[3]{x}\end{matrix}\right.\) ta có :
\(\left\{{}\begin{matrix}a+b=3\\a^2-b^3=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3-b\\\left(3-b\right)^2-b^3=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3-b\\b^3-b^2+6b-6=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=3-b\\\left(b^2+6\right)\left(b-1\right)=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{a+3}=2\\\sqrt[3]{x}=1\end{matrix}\right.\Rightarrow x=1\) ( TM )
\(\sqrt[3]{x}+\sqrt{x+3}=3\left(ĐKXĐ:x\ge-3\right)\)
\(\Rightarrow\left(\sqrt[3]{x}-1\right)+\left(\sqrt{x+3}-2\right)=0\)
\(\Rightarrow\frac{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x}^2+\sqrt[3]{x}+1\right)}{\sqrt[3]{x}^2+\sqrt[3]{x}+1}+\frac{\left(\sqrt{x+3}-2\right)\left(\sqrt{x+3}+2\right)}{\sqrt{x+3}+2}=0\)
\(\Rightarrow\frac{x-1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1}+\frac{x-1}{\sqrt{x+3}+2}=0\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1}+\frac{1}{\sqrt{x+3}+2}\right)=0\)
Có: \(\frac{1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1}+\frac{1}{\sqrt{x+3}+2}=\frac{1}{\left(\sqrt[3]{x}+\frac{1}{2}\right)^2+\frac{3}{4}}+\frac{1}{\sqrt{x+3}+2}< 0\)
Do đó: x - 1 = 0 => x = 1 (TMĐK)
Vậy phương trình có nghiệm duy nhất bằng 1
