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If $\left|\begin{array}{ll}4 & 1 \\ 2 & 1\end{array}\right|^2=\left|\begin{array}{ll}3 & 2 \\ 1 & x\end{array}\right|-\left|\begin{array}{cc}x & 3 \\ -2 & 1\end{array}\right|$, then $x=$
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6
$\begin{aligned} & \left|\begin{array}{ll}4 & 1 \\ 2 & 1\end{array}\right|\left|\begin{array}{ll}4 & 1 \\ 2 & 1\end{array}\right|=\left|\begin{array}{cc}3 & 2 \\ 1 & x\end{array}\right|-\left|\begin{array}{cc}x & 3 \\ -2 & 1\end{array}\right| \\ & \quad=\left|\begin{array}{cc}17 & 9 \\ 9 & 5\end{array}\right|=(3 x-2)-(x+6) \\ & \Rightarrow 85-81=2 x-8 \Rightarrow 4+8=2 x \Rightarrow x=6\end{aligned}$
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