2 câu trả lời
`( x - 3 )^( x + 2 ) - ( x - 3 )^( x + 8 ) = 0`
`⇒ ( x - 3 )^( x + 2 ) . 1 - ( x - 3 )^( x + 2 ) . ( x - 3 )^6 = 0`
`⇒ ( x - 3 )^( x + 2 ) . [ 1 - ( x - 3 )^6 ] = 0`
`⇒` \(\left[ \begin{array}{l}( x - 3 )^{ x + 2 } = 0\\1 - ( x - 3 )^6 = 0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}( x - 3 )^{ x + 2 } = 0^( x + 2 )\\( x - 3 )^6 = 1\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x - 3 = 0\\( x - 3 )^6 = 1^6\\ ( x - 3 )^6 = ( - 1 )^6\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x - 3 = 0\\x - 3 = 1\\ x - 3 = - 1\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x = 3\\x = 4 \\ x = 2\end{array} \right.\)
Vậy `, x ∈ { 2 ; 3 ; 4 }`
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