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In the real number system, the equation $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ has $-$
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$\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1 ; x \geq 1$
$\sqrt{(x-1)-2 \times 2 \sqrt{x-1}+4}+\sqrt{(x-1)-6 \sqrt{x-1}+9}=1$
$\begin{array}{l}
2 \sqrt{x-1}=6 \\
x=10 \\
\text { Case -II } \\
\sqrt{x-1}-2-\sqrt{x-1}+3=1 \quad 5 \leq x \leq 10
\end{array}$
Case -III
$\begin{array}{l}
-\sqrt{x-1}+2-\cdot \sqrt{x-1}+3=1 \\
2 \cdot \sqrt{x-1}=4 \\
x=5
\end{array}$
$\sqrt{(x-1)-2 \times 2 \sqrt{x-1}+4}+\sqrt{(x-1)-6 \sqrt{x-1}+9}=1$
$\begin{array}{l}
2 \sqrt{x-1}=6 \\
x=10 \\
\text { Case -II } \\
\sqrt{x-1}-2-\sqrt{x-1}+3=1 \quad 5 \leq x \leq 10
\end{array}$
Case -III
$\begin{array}{l}
-\sqrt{x-1}+2-\cdot \sqrt{x-1}+3=1 \\
2 \cdot \sqrt{x-1}=4 \\
x=5
\end{array}$
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