a, Vì \(-\dfrac{1}{2}+\sqrt{2}>0\) nên hs đồng biến trên R
\(b,x=0\Leftrightarrow y=1\\ x=1\Leftrightarrow y=-\dfrac{1}{2}+\sqrt{2}+1=\sqrt{2}+\dfrac{1}{2}\\ x=\sqrt{2}\Leftrightarrow y=-\dfrac{\sqrt{2}}{2}+1+1=2-\dfrac{\sqrt{2}}{2}\\ x=1+\sqrt{4}=3\Leftrightarrow y=-\dfrac{3}{2}+3\sqrt{2}+1=3\sqrt{2}-\dfrac{1}{2}\\ c,y=0\Leftrightarrow x=-1:\dfrac{2\sqrt{2}-1}{2}=\dfrac{2}{1-2\sqrt{2}}\\ y=1\Leftrightarrow x\left(-\dfrac{1}{2}+\sqrt{2}\right)=0\Leftrightarrow x=0\\ y=2+\sqrt{2}\Leftrightarrow x\left(-\dfrac{1}{2}+\sqrt{2}\right)=\sqrt{2}+1\\ \Leftrightarrow x=\left(\sqrt{2}+1\right):\dfrac{\left(2\sqrt{2}-1\right)}{2}=\dfrac{2\sqrt{2}+2}{2\sqrt{2}-1}\)
a: Hàm số này đồng biến trên R vì \(\sqrt{2}-\dfrac{1}{2}>0\)
