81x³ - 4x = 0

=> x.(81x² - 4) = 0

=> \(\orbr{\begin{cases}x=0\\81x^2-4=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\81x^2=4\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x^2=\frac{4}{81}\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x\in\left\{\frac{2}{9};-\frac{2}{9}\right\}\end{cases}}\)

81x^3 - 4x = 0

<=>x.(81x²-4)=0

<=>x.(9x-2)(9x+2)=0

<=>x=0 hoặc x=2/9 hoặc x=-2/9

\(81x^3-4x=0\)

\(\Rightarrow x\left(81x^2-4\right)=0\)

\(\Rightarrow x\left(9x-2\right)\left(9x+2\right)=0\)

TH1:  \(x=0\)

TH2: \(9x-2=0\Rightarrow9x=2\Rightarrow x=\frac{2}{9}\)

TH3: \(9x+2=0\Rightarrow9x=-2\Rightarrow x=-\frac{2}{9}\)

Vậy: \(x=0\) hoặc \(x=\frac{2}{9}\) hoặc \(x=-\frac{2}{9}\)

a) (x² + 4)² - 4x(x² + 4) = 0

(x² + 4)(x² + 4 - 4x) = 0

(x² + 4)(x - 2)² = 0

\(\Rightarrow\) x² + 4 = 0 hoặc (x - 2)² = 0

\(\Rightarrow\) x² = - 4 hoặc x - 2 = 0

\(\Rightarrow\) x \(\in\) tập hợp rỗng hoặc x = 2

Vậy x = 2

b) x⁵ - 18x³ + 81x = 0

x(x⁴ - 18x² + 81) = 0

x(x² - 9) = 0

x(x - 3)(x + 3) = 0

\(\Rightarrow\) x = 0 hoặc x - 3 = 0 hoặc x + 3 = 0

\(\Rightarrow\) x = 0 hoặc x = 3 hoặc x = - 3

Vậy \(x\in\left\{0;3;-3\right\}\)

\(x^5-18x^3+81x=0\)

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

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

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x^2-9=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x^2=9\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=\pm3\end{array}\right.\)

Vây ..................

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

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

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

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

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

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

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

Nhiều qua trời 

<=>\(\frac{\left(2x+1\right).\left(x+1\right)}{\left(x-3\right).\left(x+1\right)}\)+\(\frac{\left(3-2x\right).\left(x-3\right)}{\left(x+1\right).\left(x-3\right)}\)=0

<=>\(\frac{2x^2+3x+1}{\left(x-3\right).\left(x+1\right)}\)+\(\frac{3x-9-2x^2+6x}{\left(x+1\right).\left(x-3\right)}\)=0

<=>\(\frac{2x^2+3x+1+3x-9-2x^2+6x}{\left(x-3\right).\left(x+1\right)}\)=0

<=>\(\frac{12x-8}{\left(x-3\right).\left(x+1\right)}\)=0

=> 12x-8=0

=>x=2/3

3⁵ˣ⁺⁴ = 81ˣ⁺³

3⁵ˣ⁺⁴ = (3⁴)ˣ⁺³

3⁵ˣ⁺⁴ = 3⁴ˣ⁺¹²

5x + 4 = 4x +12

5x - 4x =12 - 4

x = 8