a) \(\sqrt{2x-1}\ge\sqrt{x+1}\)
⇔ 2x - 1 ≥ x + 1
⇔ x ≥ 2
Vậy x ≥ 2
b) \(\sqrt{2x}\le\sqrt{x^2}\) ( x ≥ 0)
⇔ \(\sqrt{2x}\le x\)
⇔ 2x ≤ x2
⇔ 2x - x2 ≤ 0
⇔ x.(2 - x) ≤ 0
⇔ \(\left\{{}\begin{matrix}x\le0\\2-x\le0\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x\le0\left(KTM\right)\\x\ge2\left(TM\right)\end{matrix}\right.\)
Vậy x ≥ 2
a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\2x-1>=x+1\end{matrix}\right.\Leftrightarrow x>=2\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\2x-x^2< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\left(x-2\right)\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>=2\\x=0\end{matrix}\right.\)
