a) \(3^{2x+1}+3^{2x}=36\)
\(\Leftrightarrow3^{2x}\left(3+1\right)=36\)
\(\Leftrightarrow3^{2x}.4=36\)
\(\Leftrightarrow3^{2x}=9\)
\(\Leftrightarrow3^{2x}=3^2\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
b) \(\left(x+1\right)^5=\left(x+1\right)^3\)
\(\Leftrightarrow\left(x+1\right)^5-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3\left[\left(x+1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x\in\left\{-2;0\right\}\end{cases}}\)
a)32x + 1 + 32x = 36
=> 32x . 3 + 32x = 36
=> 32x . 4 = 36
=> 32x = 36 : 4
=> 32x = 9
=> 32x = 32
=> 2x = 2
=> x = 1
b) (x + 1)5 = (x + 1)3
=> (x + 1)5 - (x + 1)3 = 0
=> (x + 1)3.[(x + 1)2 - 1] = 0
=>(x + 1)3(x + 1 - 1)(x + 1 + 1) = 0
=> (x + 1)3.x.(x + 2) = 0
=> x = -1 hoặc x = 0 hoặc x = -2
