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) \(x^5-18x^3+81x=0\)

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

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

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

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

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

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

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

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

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

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

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

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

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

  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

12?

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

b: \(\begin{cases}81x^3y^2-81x^2y^2+33xy^2-29y^2=4\left(1\right)\\ 25y^3+9x^2y^3-6xy^3-4y^2=24\left(2\right)\end{cases}\)

Khi y=0 thì (2) sẽ trở thành: \(25\cdot0^3+9x^2\cdot0^3-6\cdot x\cdot0^3-4\cdot0^2=24\)

=>0=24(loại)

=>y<>0

Ta có: \(81x^3y^2-81x^2y^2+33xy^2-29y^2=4\)

=>\(y^2(81x^3-81x^2+33x-29)=4\)

=>\(81x^3-81x^2+33x-29=\frac{4}{y^2}\)

Ta có: \(25y^3+9x^2y^3-6xy^3-4y^2=24\)

=>\(y^3(9x^2 - 6x + 25) - 4y^2 = 24\)

=>\(y(9x^2 - 6x + 25) - 4 = \frac{24}{y^2}\)

Đặt u=3x-1

=>3x=u+1

=>\(x=\frac{u+1}{3}\)

\(9x^2-6x+25=9x^2-6x+1+24=\left(3x-1\right)^2+24\)

\(=u^2+24\)

(1) sẽ tương đương: \(81\left(\frac{u+1}{3}\right)^3 - 81\left(\frac{u+1}{3}\right)^2 + 33\left(\frac{u+1}{3}\right) - 29 = \frac{4}{y^2}\)

=>\(3(u+1)^3 - 9(u+1)^2 + 11(u+1) - 29 = \frac{4}{y^2}\)

=>\(3u^3+2u-24=\frac{4}{y^2}\quad(3)\)

THay (3) vào phương trình (2), ta được:

\(y(u^2+24)-4=\frac{24}{y^2}\quad(4)\)

Ta có: \(3u^3+2u-24=\frac{4}{y^2}\quad(3)\)

=>\(18u^3+12u-144=\frac{24}{y^2}\)

=>\(y(u^2 + 24) - 4 = 18u^3 + 12u - 144\)

=>\(y(u^2+24)=18u^3+12u-140\)

=>\(y=\frac{18u^3 + 12u - 140}{u^2 + 24}\)

Thay lại vào (3), ta được: \(\frac{4}{\left(\frac{18u^3 + 12u - 140}{u^2 + 24}\right)^2} = 3u^3 + 2u - 24\)

=>\((9u^3 + 6u - 70)^2(3u^3 + 2u - 24) = (u^2 + 24)^2\)

=>u=2

\(y=\frac{18\cdot2^3+12\cdot2-140}{2^2+24}=\frac{144 + 24 - 140}{28}=\frac{28}{28}=1\)

3x-1=2

=>3x=3

=>x=1