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

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

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

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

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

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

  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

Nhiều qua trời 

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^3 + 81x - 170 = 0`

`<=>x^3 - 2x^2 + 2x^2 - 4x + 85x - 170 = 0`

`<=> x^2 ( x - 2 ) + 2x ( x - 2 ) + 85 ( x - 2 ) = 0`

`<=> ( x - 2 ) ( x^2 + 2x + 85 ) = 0`

`<=> ( x - 2 ) [ ( x + 1 )^2 + 84 ] = 0`

  Mà `( x + 1 )^2 + 84 > 0 AA x`

   `=> x - 2 = 0`

`<=> x = 2`

Vậy `S = { 2 }`

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

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

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

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

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

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

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

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

Vây ..................

`#3107`

\(3^{8x+4}=81^{x+3}\\ \Rightarrow3^{8x+4}=\left(3^4\right)^{x+3}\\ \Rightarrow3^{8x+4}=3^{4x+12}\\ \Rightarrow8x+4=4x+12\\ \Rightarrow8x-4x=12-4\\ \Rightarrow4x=8\\ \Rightarrow x=8\div4\\ \Rightarrow x=2\\ \text{Vậy, x = 2.}\)

Lời giải:

$3^{8x+4}=81^{x+3}$

$3^{8x+4}=(3^4)^{x+3}$

$3^{8x+4}=3^{4(x+3)}$

$\Rightarrow 8x+4=4(x+3)$
$\Rightarrow 2x+1=x+3$

$\Rightarrow x=2$

a, Đặt \(\sqrt[3]{81x-8}=3y-2\Leftrightarrow9x=3y^3-6y^2+4y\left(1\right)\)

Phương trình tương đương: \(3y-2=x^3-2x^2+\dfrac{4}{3}x-2\)

\(\Leftrightarrow9y=3x^3-6x^2+4x\)

Ta có hệ: \(\left\{{}\begin{matrix}9x=3y^3-6y^2+4y\\9y=3x^3-6x^2+4x\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)\left(3x^2+3y^2+3xy-6x-6y+13\right)=0\)

Vì \(3x^2+3y^2+3xy-6x-6y+13\)

\(=\dfrac{1}{2}\left[3\left(x+y\right)^2+3\left(x-2\right)^2+3\left(y-2\right)^2+2\right]>0\) nên \(x=y\)

Khi đó: \(\left(1\right)\Leftrightarrow3x^3-6x^2-5x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3\pm2\sqrt{6}}{3}\end{matrix}\right.\)

Thử lại ta được \(x=0;x=\dfrac{3\pm2\sqrt{6}}{3}\) là các nghiệm của phương trình.

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