tìm x biết: $(x-3)^{x+2}$ – $(x-3)^{x+8}$ =0

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tìm x biết: $(x-3)^{x+2}$ – $(x-3)^{x+8}$ =0

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Mít Mít 2 years 2021-05-10T05:36:20+00:00 2 Answers 10 views 0

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    0
    2021-05-10T05:37:21+00:00

    Đáp án:

    $x=2,\ x=3$ hoặc $x=4$ 

    Giải thích các bước giải:

    \(\begin{array}{l}
    \quad (x-3)^{x+2} – (x-3)^{x+8}=0\\
    \Leftrightarrow (x-3)^{x+2} – (x-3)^{x+2}.(x-3)^6=0\\
    \Leftrightarrow (x-3)^{x+2}\left[1 – (x-3)^6\right]=0\\
    \Leftrightarrow \left[\begin{array}{l}(x-3)^{x+2} =0\\1 – (x-3)^6 =0\end{array}\right.\\
    \Leftrightarrow \left[\begin{array}{l}(x-3)^{x+2} =0\\(x-3)^6 =1\end{array}\right.\\
    \Leftrightarrow \left[\begin{array}{l}x-3 =0\\x-3 =1\\x-3=-1\end{array}\right.\\
    \Leftrightarrow \left[\begin{array}{l}x=3\\x=4\\x=2\end{array}\right.\\
    \text{Vậy $x=2,\ x=3$ hoặc $x=4$}
    \end{array}\) 

    0
    2021-05-10T05:37:38+00:00

    `(x-3)^(x+2)-(x-3)^(x+8)=0`

    ⇔`(x-3)^(x+2)*[1-(x-3)^6]=0`

    Th1:

    `(x-3)^(x+2)=0⇔x=3`

    Th2:

    `1-(x-3)^6=0`

          ⇔`1=(x-3)^6`

          ⇔$\left \{ {{x-3=-1} \atop {x-3=1}} \right.$

          ⇔$\left \{ {{x=2} \atop {x=4}} \right.$ 

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