a) \(\left(x^2+4\right)^2-4x\left(x^2+4\right)=0\)

\(=\left(x^2+4\right)\left(x^2+4-4x\right)=0\)

\(=\left(x^2+4\right)\left(x+2\right)^2=0\)

Mà \(x^2\ge0\Rightarrow x^2+4>0\)

\(\Rightarrow x+2=0\)

\(\Rightarrow x=-2\)

b) \(x^5-18x^3+81x=0\)

\(=\left(x^5-9x^3\right)-\left(9x^3-81x\right)=0\)

\(=x^3\left(x^2-9\right)-9x\left(x^2-9\right)=0\)

\(=\left(x^3-9x\right)\left(x^2-9\right)=0\)

\(=x\left(x^2-9\right)\left(x^2-9\right)=0\)

\(=x\left(x^2-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x\in\left\{-3;3\right\}\end{cases}}\)

\(3^{8x+4}=81^{x+3}\)

\(3^{8x+4}=\left(3^4\right)^{x+3}\)

\(3^{8x+4}=3^{4x+12}\)

\(\Rightarrow8x+4=4x+12\)

\(\Rightarrow8x-4x=12-4\)

\(\Rightarrow4x=8\Rightarrow x=2\)

2

  3^{8.x + 4} = 81^{x + 3}

  3^{8.x + 4} = (3⁴)^{x + 3}

  3^{8.x + 4} = 3^{4.x + 12}

  8.x + 4 = 4.x + 12

8.x - 4.x = 12 - 4

        4.x = 8

           x = 8 : 4

           x = 2

3⁵ˣ⁺⁴ = 81ˣ⁺³

3⁵ˣ⁺⁴ = (3⁴)ˣ⁺³

3⁵ˣ⁺⁴ = 3⁴ˣ⁺¹²

5x + 4 = 4x +12

5x - 4x =12 - 4

x = 8

d,Sửa đề

\(-x^3+6x^2-12x+8\)

\(=-\left(x^3-6x^2+12x-8\right)\)

\(=-\left(x^3-3.x^2.2+3.x.2^2-2^3\right)\)

\(=-\left(x-2\right)^3\)

\(e,27x^3+81x^2+81x+27\)

\(=27\left(x^3+3x^2+3x+1\right)\)

\(=27\left(x+1\right)^3\)

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 \(T = \left\{ {0;\dfrac{{3 \pm 2\sqrt 6 }}{3}} \right\}\)