Question

Answer 1

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

Answer 2

a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\2x-1>=x+1\end{matrix}\right.\Leftrightarrow x>=2\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\2x-x^2< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\left(x-2\right)\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>=2\\x=0\end{matrix}\right.\)