\(\left|2x-1\right|+\left(\dfrac{2}{3}-x\right)^{2024}=0\)
\(\left|2x-1\right|=-\left(\dfrac{2}{3}-x\right)^{2024}\)
Vì \(VT\ge0;VP\le0\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}2x-1=0\\\dfrac{2}{3}-x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)(Loại)
\(\left|2x-1\right|+\left(\dfrac{2}{3}-x\right)^{2024}=0\)
Nhận xét: +) \(\left|2x-1\right|\ge0,\forall x\)
\(\left(\dfrac{2}{3}-x\right)^{2024}\ge0,\forall x\)
\(\Rightarrow\left|2x-1\right|+\left(\dfrac{2}{3}-x\right)^{2024}\ge0,\forall x\)
Do đó, \(\left|2x-1\right|+\left(\dfrac{2}{3}-x\right)^{2024}=0\) khi:
\(\left\{{}\begin{matrix}2x-1=0\\\dfrac{2}{3}-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\Rightarrow x\in\varnothing}\)
Vậy \(x\in\varnothing\)
Phần bị lỗi là:
\(\left\{{}\begin{matrix}2x-1=0\\\dfrac{2}{3}-x=0\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow x\in\varnothing\)
Bạn thông cảm nhé!