ĐKXĐ: \(x\geq -5\).
\(PT\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\left(\sqrt{x+5}+\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\sqrt{x+5}+\dfrac{1}{2}\left(1\right)\\\dfrac{3}{2}-x=\sqrt{x+5}+\dfrac{1}{2}\left(2\right)\end{matrix}\right.\).
+) \(\left(1\right)\Leftrightarrow x-2=\sqrt{x+5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x+4=x+5\end{matrix}\right.\Leftrightarrow x=\dfrac{5+\sqrt{29}}{2}\).
+) \(\left(2\right)\Leftrightarrow\sqrt{x+5}=1-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x+5=x^2-2x+1\end{matrix}\right.\Leftrightarrow x=-1\).
Do đó hai nghiệm của pt này có tổng bằng \(\dfrac{3+\sqrt{29}}{2}\). Chọn A
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