a.
Đặt \(\left(2+\sqrt{3}\right)^x=t>0\Rightarrow t^2=\left(7+4\sqrt{3}\right)^x\)
Phương trình trở thành:
\(t^2+2t-3=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-3< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left(2+\sqrt{3}\right)^x=1\)
\(\Rightarrow x=0\)
b.
\(2^{x^2+x}-4.2^{x^2-x}-2^{2x}+4=0\)
\(\Leftrightarrow2^{x^2-x}\left(2^{2x}-4\right)-\left(2^{2x}-4\right)=0\)
\(\Leftrightarrow\left(2^{x^2-x}-1\right)\left(2^{2x}-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2^{x^2-x}-1=0\\2^{2x}-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=0\\2x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
