Toán tìm x biết: $(x-3)^{x+2}$ – $(x-3)^{x+8}$ =0 08/09/2021 By Eloise tìm x biết: $(x-3)^{x+2}$ – $(x-3)^{x+8}$ =0
Đáp án: $x=2,\ x=3$ hoặc $x=4$ Giải thích các bước giải: \(\begin{array}{l}\quad (x-3)^{x+2} – (x-3)^{x+8}=0\\\Leftrightarrow (x-3)^{x+2} – (x-3)^{x+2}.(x-3)^6=0\\\Leftrightarrow (x-3)^{x+2}\left[1 – (x-3)^6\right]=0\\\Leftrightarrow \left[\begin{array}{l}(x-3)^{x+2} =0\\1 – (x-3)^6 =0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{l}(x-3)^{x+2} =0\\(x-3)^6 =1\end{array}\right.\\\Leftrightarrow \left[\begin{array}{l}x-3 =0\\x-3 =1\\x-3=-1\end{array}\right.\\\Leftrightarrow \left[\begin{array}{l}x=3\\x=4\\x=2\end{array}\right.\\\text{Vậy $x=2,\ x=3$ hoặc $x=4$}\end{array}\) Trả lời
`(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` ⇔$\left \{ {{x-3=-1} \atop {x-3=1}} \right.$ ⇔$\left \{ {{x=2} \atop {x=4}} \right.$ Trả lời
Đáp án:
$x=2,\ x=3$ hoặc $x=4$
Giải thích các bước giải:
\(\begin{array}{l}
\quad (x-3)^{x+2} – (x-3)^{x+8}=0\\
\Leftrightarrow (x-3)^{x+2} – (x-3)^{x+2}.(x-3)^6=0\\
\Leftrightarrow (x-3)^{x+2}\left[1 – (x-3)^6\right]=0\\
\Leftrightarrow \left[\begin{array}{l}(x-3)^{x+2} =0\\1 – (x-3)^6 =0\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}(x-3)^{x+2} =0\\(x-3)^6 =1\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x-3 =0\\x-3 =1\\x-3=-1\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x=3\\x=4\\x=2\end{array}\right.\\
\text{Vậy $x=2,\ x=3$ hoặc $x=4$}
\end{array}\)
`(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`
⇔$\left \{ {{x-3=-1} \atop {x-3=1}} \right.$
⇔$\left \{ {{x=2} \atop {x=4}} \right.$