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If $\left|\begin{array}{ccc}x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4\end{array}\right|=0$, then $x=$
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The correct answer is:
$1,-9$
By $C_1 \rightarrow C_1+C_2+C_3$,
we have $(9+x)$ $\left|\begin{array}{ccc}1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4\end{array}\right|=0$
$\Rightarrow (x+9)$ $\left|\begin{array}{ccc}1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4\end{array}\right|=0$
\(\begin{aligned} & \Rightarrow(x+9)\left|\begin{array}{ccc}0 & 1-x & 0 \\ 0 & x-1 & 1-x \\ 1 & 3 & x+4\end{array}\right|=0 \\ & \Rightarrow(x+9) \cdot(1-x)^2=0 \\ & \Rightarrow \text { So, } x=1,1,-9 .\end{aligned}\)
we have $(9+x)$ $\left|\begin{array}{ccc}1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4\end{array}\right|=0$
$\Rightarrow (x+9)$ $\left|\begin{array}{ccc}1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4\end{array}\right|=0$
\(\begin{aligned} & \Rightarrow(x+9)\left|\begin{array}{ccc}0 & 1-x & 0 \\ 0 & x-1 & 1-x \\ 1 & 3 & x+4\end{array}\right|=0 \\ & \Rightarrow(x+9) \cdot(1-x)^2=0 \\ & \Rightarrow \text { So, } x=1,1,-9 .\end{aligned}\)
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