Question

tìm x biết |2x-1|+(2/3-x)^2024=0

Answer 1

\(\left|2x-1\right|+\left(\dfrac{2}{3}-x\right)^{2024}=0\)

\(\left|2x-1\right|=-\left(\dfrac{2}{3}-x\right)^{2024}\)

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)

Answer 2

Tính Vẫn Bằng 0 Em Nhé!

Answer 3

\(\left|2x-1\right|+\left(\dfrac{2}{3}-x\right)^{2024}=0\)

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 \(x\in\varnothing\)

Answer 4

Phần bị lỗi là:
\(\left\{{}\begin{matrix}2x-1=0\\\dfrac{2}{3}-x=0\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)

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