\(3^{x+1}=9^x\)
\(\Rightarrow3^{x+1}=3^{2x}\)
\(\Rightarrow3^{2x}-3^{x+1}=0\)
\(\Rightarrow3^x.\left(3^x-3\right)=0\)
Vì \(3^x>0\) \(\forall x.\)
\(\Rightarrow3^x-3=0\)
\(\Rightarrow3^x=0+3\)
\(\Rightarrow3^x=3\)
\(\Rightarrow3^x=3^1\)
\(\Rightarrow x=1\)
Vậy \(x=1.\)
Chúc bạn học tốt!
\(3^{x+1}\) = \(9^x\)
\(3^{x+1}\) = \(3^2\)
x + 1 = 2
x = 2-1
x = ?
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