\(9^n-2\cdot3^n+x^2+5+4x=0\)
\(\Leftrightarrow\left(9^n-2\cdot3^n+1\right)+\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left[\left(3^n\right)^2-2\cdot3^n\cdot1+1^2\right]+\left(x^2+2\cdot x\cdot2+2^2\right)=0\)
\(\Leftrightarrow\left(3^n-1\right)^2+\left(x+2\right)^2=0\)
Ta thấy: \(\left(3^n-1\right)^2\ge0\forall n\)
\(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(3^n-1\right)^2+\left(x+2\right)^2\ge0\forall x;n\)
Mặt khác: \(\left(3^n-1\right)^2+\left(x+2\right)^2=0\)
nên ta có: \(\left\{{}\begin{matrix}3^n-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3^n=1\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=0\\x=-2\end{matrix}\right.\left(tm\right)\)
Vậy \(n=0;x=-2\).
#\(Toru\)
