Lời giải:
PT $\Leftrightarrow 27\sqrt[3]{81x-8}=27x^3-54x^2+36x-54$
$\Leftrightarrow 27\sqrt[3]{81x-8}=(3x-2)^3-46$
Đặt $\sqrt[3]{81x-8}=a; 3x-2=b$. Khi đó:
\(\left\{\begin{matrix} a^3-27b=46\\ 27a=b^3-46\end{matrix}\right.\) $\Rightarrow 27a=b^3-(a^3-27b)$
$\Leftrightarrow a^3-b^3+27a-27b=0$
$\Leftrightarrow (a-b)(a^2+ab+b^2+27)=0$
Dễ thấy $a^2+ab+b^2+27>0$ với mọi $a,b\in\mathbb{R}$
Do đó $a-b=0\Rightarrow a=b$
$\Leftrightarrow 81x-8=(3x-2)^3$
$\Leftrightarrow 27x^3-54x^2-45x=0$
$\Rightarrow x=0; x=\frac{3\pm 2\sqrt{6}}{3}$
Vậy.......
\(\sqrt[3]{{81x - 8}} = {x^3} - 2{x^2} + \dfrac{4}{3}x - 2\left( 1 \right)\)
\(\left( 1 \right) \Leftrightarrow 27{x^3} - 54{x^2} + 36x - 54 = 27\sqrt[3]{{81x - 8}} \)
Đặt \(y=\sqrt[3]{81x-8}\Leftrightarrow y^3=81x-8\)
Vậy ta có hệ phương trình \(\left\{{}\begin{matrix}27x^3-54x^2+36x-54=27y\\81x-8=y^3\end{matrix}\right.\Rightarrow\left(3x-2\right)^3+27\left(3x-2\right)=y^3+y\left(2\right)\)
Xét hàm số \(f(t)=t^3+t(t \in \mathbb{R})\)
Đạo hàm \(f'\left(t\right)=3t^2+1>0;\forall t\in\) \(\mathbb{R}\)
Vậy hàm số trên đồng biến trên \(\mathbb{R}\)
\(\left(2\right)\Leftrightarrow f\left(3x-2\right)=f\left(y\right)\\ \Leftrightarrow3x-2=y\\ \Leftrightarrow3x-2=\sqrt[3]{81x-8}\\ \Leftrightarrow27x^3-54x^2-45x=0\)
\(\Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{{3 \pm 2\sqrt 6 }}{3} \end{array} \right.\)
Vậy phương trình có tập nghiệm: \(T = \left\{ {0;\dfrac{{3 \pm 2\sqrt 6 }}{3}} \right\}\)
Cách khác:
Phương trình đã cho tương đương với \(3.\sqrt[3]{{3\left( {x - \dfrac{2}{3}} \right) + \dfrac{{46}}{{27}}}} = {\left( {x - \dfrac{2}{3}} \right)^2} - \dfrac{{46}}{{27}}\)
Đặt \(\left\{ \begin{array}{l} u = x - \dfrac{2}{3}\\ v = \sqrt[3]{{3\left( {x - \dfrac{2}{3}} \right) + \dfrac{{46}}{{27}}}} = \sqrt[3]{{3u + \dfrac{{46}}{{27}}}} \end{array} \right.\) ta có hệ: \(\left\{ \begin{array}{l} 3u = {v^3} - \dfrac{{46}}{{27}}\\ 3v = {u^3} - \dfrac{{46}}{{27}} \end{array} \right. \)
Trừ hai phương trình cho nhau theo từng vế ta có:
\(3\left( {u - v} \right) = \left( {v - u} \right)\left( {{v^2} + uv + {u^2}} \right) \Leftrightarrow \left[ \begin{array}{l} u - v = 0{\rm{ }}\left( 1 \right)\\ {v^2} + uv + {u^2} = - 3{\rm{ }}\left( 2 \right) \end{array} \right. \)
Dễ thấy \(v^2+uv+u^2\ge0\) nên \((2)\) vô nghiệm.
\(\left( 1 \right) \Leftrightarrow u = v \Rightarrow \sqrt[3]{{3x - \dfrac{8}{{27}}}} = x - \dfrac{2}{3} \Leftrightarrow {x^3} - 2{x^2} - \dfrac{5}{3} = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{{3 \pm 2\sqrt 6 }}{3} \end{array} \right.\)
Vậy \(T = \left\{ {0;\dfrac{{3 \pm 2\sqrt 6 }}{3}} \right\}\)
Cách khác:
\(\sqrt[3]{81x-8}=x^3-2x^2+\frac{4}{3}x-2\)
<=> \(3\sqrt[3]{81x-8}=3x^3-6x^2+4x-6\)
<=>\(3\left[\sqrt[3]{81x-8}-\left(3x-2\right)\right]=3x^3-6x^2-5x\)
Xét \(\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{81x-8}=a\\3x-2=b\end{matrix}\right.\)
Có \(a^2+ab+b^2=0\) <=> \(\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}=0\)
Vì \(VT\ge0\).Dấu "=" xảy ra <=>\(\left\{{}\begin{matrix}b=0\\a+\frac{b}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=0\\a=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\\sqrt[3]{81x-8}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2}{3}\\x=\frac{8}{81}\end{matrix}\right.\)(L)
=> \(\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2\ne0\)
pt <=>\(\frac{81x-8-\left(3x-2\right)^3}{\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2}=3x^3-6x^2-5x\)
<=>\(\frac{-27x^3+54x^2+45x}{\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2}-\left(3x^3-6x^2-5x\right)=0\)
<=>\(\left(3x^3-6x^2-5x\right)\left(\frac{-9}{\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2}-1\right)=0\)
<=>\(\left[{}\begin{matrix}3x^3-6x^2-5x=0\\-\frac{9}{\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2}=1\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x\left(x-\frac{3+2\sqrt{6}}{3}\right)\left(x-\frac{3-2\sqrt{6}}{3}\right)=0\left(1\right)\\-9=\sqrt[3]{\left(81-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2\left(2\right)\end{matrix}\right.\)
Pt (1) <=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=\frac{3+2\sqrt{6}}{3}\left(tm\right)\\x=\frac{3-2\sqrt{6}}{3}\left(tm\right)\end{matrix}\right.\)
Pt (2) vô nghiệm do VP >0
Vậy pt có tập nghiệm \(S=\left\{0,\frac{3+2\sqrt{6}}{3},\frac{3-2\sqrt{6}}{3}\right\}\)
