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Đặt
$ \begin{array}{l}
u = a - x\\
v = x - b\\
\Leftrightarrow \left\{ \begin{array}{l}
u + v = a - b\\
{u^5} + {v^5} = {\left( {a - b} \right)^5}
\end{array} \right.\,\,\,\,\,\,\,(*)
\end{array} $
Ta có :
$ \begin{array}{l}
{u^5} + {v^5} = \left( {u + v} \right)\left\{ {\left[ {{{\left( {u + v} \right)}^2} - 2uv} \right] - uv{{\left( {u + v} \right)}^2} + {u^2}{v^2}} \right\}\\
\Rightarrow {\left( {a - b} \right)^5} = \left( {a - b} \right)\left\{ {{{\left[ {{{\left( {a - b} \right)}^2} - 2uv} \right]}^2} - uv{{\left( {a - b} \right)}^2} + {u^2}{v^2}} \right\}\\
\Leftrightarrow {\left( {a - b} \right)^4} = {\left( {a - b} \right)^4} - 4{\left( {a - b} \right)^2}uv + 4{\left( {uv} \right)^2} - uv{\left( {a - b} \right)^2} + {\left( {uv} \right)^2}\\
\Leftrightarrow 5{\left( {uv} \right)^2} - 5{\left( {a - b} \right)^2}\left( {uv} \right) = 0\\
\Leftrightarrow uv = 0\,\,\,\,\,\, \vee \,\,\,\,\,uv = {\left( {a - b} \right)^2}
\end{array} $
Do đó ta có:
$ \begin{array}{l}
(*) \Leftrightarrow \left\{ \begin{array}{l}
u + v = a - b\\
uv = 0
\end{array} \right.\,\,\,\,\,\,\,\,(1) \vee \,\,\,\left\{ \begin{array}{l}
u + v = a - b\\
uv = {\left( {a - b} \right)^2}
\end{array} \right.\,\,\,\,\,\,\,\,\,(2)\\
(1) \Leftrightarrow \left\{ \begin{array}{l}
u = 0\\
v = a - b
\end{array} \right.\,\,\,\,\,\, \vee \left\{ \begin{array}{l}
u = a - b\\
v = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a - x = 0\\
x - b = a - b
\end{array} \right.\,\,\,\,\,\, \vee \left\{ \begin{array}{l}
a - x = a - b\\
x - b = 0
\end{array} \right.\\
\Leftrightarrow x = a\,\,\,\,\,\, \vee \,\,\,\,x = b
\end{array} $
(2) vô nghiệm
Vậy phương trình đã cho có 2 nghiệm là : $ {x_1} = a;\,\,\,\,{x_2} = b $
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