ta có 81=3^4

=> pt trở thành

3^4x+4=3^4x+3

3^4x-3^4x=3-4

0x=-1

=>pt vô nghiệm

\(^{\left(3^4\right)^{x+1}=81^{x+3}\Rightarrow81^{x+1}=81^{x+3}\Rightarrow x+1=x+3}\)vô nghiệm vì 1 khác 3

3^4x+4 = 81^x+3

3^4x+4 = 3^4(x+3)

3^4x+4 = 3^4x+12

<=> 4x+4 = 4x+12

4x-4x = 12-4

0 = 8

Vậy không có x thỏa mãn

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

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

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

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

=>4x+4=4x+12

=>4x-4x+4-12=0

=>0-8=0

=>-8=0 (vô lí)

Vậy không tồn tại x thỏa mãn.

chịu nè!!!

1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)

\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)

\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)

\(\Leftrightarrow5x-6=0\)

\(\Leftrightarrow5x=6\)

\(\Leftrightarrow x=\frac{6}{5}\)

Vậy: \(x=\frac{6}{5}\)

2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)

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

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)

\(\Leftrightarrow12x+24=0\)

\(\Leftrightarrow12x=-24\)

\(\Leftrightarrow x=-2\)

Vậy: x=-2

3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)

\(\Leftrightarrow15x-30=0\)

\(\Leftrightarrow15x=30\)

\(\Leftrightarrow x=2\)

Vậy: x=2

4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)

\(\Leftrightarrow83x-83=0\)

\(\Leftrightarrow83x=83\)

\(\Leftrightarrow x=1\)

Vậy: x=1

1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)

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

2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)

3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)

4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)

\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)

5, em xem lại đề nhé

à lag tý @@

5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)

a)\(x^3+4x^2+4x=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

b)\(\left(x+3\right)^2-4=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3-2=0\\x+3+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}}\)

c)\(x^4-9x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

d)\(x^2-6x+9=81\)

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)

e)\(x^3+6x^2+9x-4x=0\)

\(\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0;x=-5\\x=-1\end{cases}}}\)

#H

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

\(\Leftrightarrow3^{4x+4}=3^{4x+16}\)

\(\Rightarrow\) không tồn tại x thỏa mãn      

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

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

=>\(3^{4x+4}=3^{4x+16}\)

=>4x+4=4x+16

=>4=16 (vô lí)

Vậy không tồn tại x thỏa mãn .

please help me tick!!!

lam on cho minh vai ****