\(3^{x+2}+3^{x+1}-3^x=33\)
=>\(3^x\cdot9+3^x\cdot3-3^x=33\)
=>\(3^x\cdot11=33\)
=>\(3^x=3\)
=>x=1
\(3^{x+2}+3^{x+1}-3^x=33\\ 3^x.3^2+3^x.3-3^x=33\\ 3^x.\left(9+3-1\right)=33\\ 3^x.11=33\\ 3^x=33:11\\ 3^x=3\\ \Rightarrow x=1\)
\(a,3^{x+2}+3^{x+1}-3^x=33\)
\(3^x.9+3^x.3-3^x.1=33\)
\(3^x.\left(9+3-1\right)=33\)
\(3^x.11=33\)
\(3^x=33:11\)
\(3^x=3\)
\(=>x=1\)
Vậy `x = 1`