Lời giải:
\(x^3+6x^2+12x+6=3\sqrt[3]{3x+8}\)
\(\Leftrightarrow x^3+6x^2+12x=3(\sqrt[3]{3x+8}-2)\)
\(\Leftrightarrow x(x^2+6x+12)=\frac{3.3x}{\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4}\)
\(\Leftrightarrow x\left[(x^2+6x+12)-\frac{9}{\sqrt[3]{(3x+8)^2+2\sqrt[3]{3x+8}+4}}\right]=0\)
TH1: \(x=0\) (thỏa mãn)
TH2: Biểu thức trong ngoặc vuông bằng 0
Ta thấy \(x^2+6x+12=(x+3)^2+3\geq 3\forall x\in\mathbb{R}\) (1)
\(\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4=(\sqrt[3]{3x+8}+1)^2+3\geq 3\)
\(\Rightarrow \frac{9}{\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4}\leq 3\) (2)
Từ (1), (2) suy ra \(x^2+6x+12-\frac{9}{\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4}\geq 0\)
Dấu bằng xảy ra khi \(x^2+6x+12=\frac{9}{\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4}=3\Leftrightarrow \left\{\begin{matrix} (x+3)^2=0\\ (\sqrt[3]{3x+8}+1)^2=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x=-3\\ x=-3\end{matrix}\right.\) (thỏa mãn)
Vậy \(x\in\left\{-3;0\right\}\)
✿ Another way ✿
\(x^3+6x^2+12x+6=3\sqrt[3]{3x+8}\)
\(\Leftrightarrow\left(x+2\right)^3-3\sqrt[3]{3x+8}=2\)
☘ Đặt \(x+2=a\text{ và }\sqrt[3]{3x+8}=b\)
\(\Rightarrow\left\{{}\begin{matrix}a^3-3b=2\\b^3-3a=2\end{matrix}\right.\)
☘ Trừ vế theo vế
\(\Rightarrow a^3-b^3-3b+3a=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+3\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3\right)=0\)
☘ Suy ra a = b
\(\Leftrightarrow x+2=\sqrt[3]{3x+8}\)
\(\Leftrightarrow x^3+6x^2+12x+8=3x+8\)
\(\Leftrightarrow x\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\left(\text{nhận}\right)\)
⚠ Tự kết luận nha.
