=243x2+4x243+486x2-243

=2x243+5x243+2x2x243-243

=7x243+4x243-1x243

=11x243-1x243

=10x243

=2430

=>  (x-1)^5 = (-3)^5

=>  x-1 = -3

=>  x = -2

(x-1)^5=-243

(x-1)^5=(-3)^5

x-1    =-3

  x    =(-3)+1

  x    = -2

\(\left(x-1\right)^5=-243\)

\(\left(x-1\right)^5=-243\)

\(\left(x-1\right)^5=\left(-3\right)^5\)

\(x-1=-3\)

\(x=-3+1\)

\(x=-2\)

cứu

a: =>(2x-1)^5=2^5

=>2x-1=2

=>2x=3

=>x=3/2

b: =>3^x=243*27=3^5*3^3=3^8

=>x=8

\(a,\left(2x-1\right)^5=32\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\\ b,\dfrac{1}{27}\cdot3^x=243\\ \Leftrightarrow3^x=6561\\ \Leftrightarrow3^x=3^8\\ \Leftrightarrow x=8\)

a: =>1/3x-2/5x=5

=>-1/15x=5

=>x=-75
b: =>4x=4

=>x=1

c: =>6*3^x-5*3^x=243

=>3^x=243

=>x=5

a) \(3^{2x-1}=243\)

\(\Leftrightarrow3^{2x-1}=3^5\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=5+1\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=\dfrac{6}{2}\)

\(\Leftrightarrow x=3\)

Vậy \(x=3\)

b) \(\left(3^x\right)^2:3^3=\dfrac{1}{243}\)

\(\Leftrightarrow3^{2x}:3^3=\dfrac{1}{3^5}\)

\(\Leftrightarrow3^{2x}:3^3=3^{-5}\)

\(\Leftrightarrow3^{2x-3}=3^{-5}\)

\(\Leftrightarrow2x-3=-5\)

\(\Leftrightarrow2x=-5+3\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{2}\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

c) \(2^{3x+2}=4^{x+5}\)

\(\Leftrightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)

\(\Leftrightarrow2^{3x+2}=2^{2\left(x+5\right)}\)

\(\Leftrightarrow3x+2=2\left(x+5\right)\)

\(\Leftrightarrow3x+2=2x+10\)

\(\Leftrightarrow3x-2x=10-2\)

\(\Leftrightarrow x=8\)

Vậy \(x=8\)

d) \(3^{x+1}=9^x\)

\(\Leftrightarrow3^{x+1}=\left(3^2\right)^x\)

\(\Leftrightarrow3^{x+1}=3^{2x}\)

\(\Leftrightarrow x+1=2x\)

\(\Leftrightarrow2x-x=1\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

a: =>4^x=640

=>\(x\in\varnothing\)

b: =>\(3^{-2x}\cdot3^{3x}=243\)

=>3^x=243

=>x=5

a) \(\left(2x-3\right)^2=16\)

\(\left(2x-3\right)^2=4^2\)

\(2x-3=4\)

\(2x=7\)

\(x=\dfrac{7}{2}=3,5\)

b) \(\left(3x-2\right)^5=-243\)

\(\left(3x-2\right)^5=-3^5\)

\(3x-2=-3\)

\(3x=-1\)

\(3x=-\dfrac{1}{3}\)

c) \(\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\left(x-7\right)^{x+1}\times\left[1-\left(x-7\right)^{10}\right]=0\)

\(\left(x-7\right)^{x+1}=0\) ; \(1-\left(x-7\right)^{10}=0\)

\(x-7=0;\left(x-7\right)^{10}=1\)

\(x=7;\left(x-7=1;x-7=-1\right)\)

\(x=7;x=8;x=6\)

a, (2\(x\) - 3)² = 16

     \(\left[{}\begin{matrix}2x-3=-4\\2x-3=4\end{matrix}\right.\)

     \(\left[{}\begin{matrix}2x=-1\\2x=7\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(x\in\){ - \(\dfrac{1}{2}\); \(\dfrac{7}{2}\)}

b, (3\(x\) - 2)⁵   = -243 

    ( 3\(x\) - 2)⁵ =  (-3)⁵

 3\(x\)    -  2     = -3

     3 \(x\)           = -1

        \(x\)          = - \(\dfrac{1}{3}\)

Vậy \(x\) = -\(\dfrac{1}{3}\)

c, \(\left(x-7\right)\)^\(x+1\)  = (\(x-7\))^\(x+11\)

    (\(x-7\))^{\(^{x+1}\)}.( \(\left(x-7\right)^{10}\) -  1 ) = 0

      \(\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\)

         ^{\(\left[{}\begin{matrix}x=7\\x-7=-1\\x-7=1\end{matrix}\right.\)}

         \(\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)

Vậy \(x\in\){ 6; 7; 8}

Câu a mình thiếu 2 trường hợp 

\(\left(2x-3\right)^2=16\)

\(\left(2x-3\right)^{ 2}=4^2\)

\(2x-3=4;2x-3=-4\)

\(2x=7;2x=-1\)

\(x=\dfrac{7}{2};x=-\dfrac{1}{2}\)

\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)

\(\)\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\left(\dfrac{1}{3}\right)^5\)

⇒\(\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\\\dfrac{2}{3}x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{2}{3}\\\dfrac{2}{3}x=0\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)