[TEX]DK: x \geq \frac{1}{2}[/TEX]
[TEX]PT \Leftrightarrow f(x)= \sqrt{1+log_2x}+log_2x=2m[/TEX]
[TEX]f(x) \geq -1 \forall x \geq \frac{1}{2}[/TEX]
[TEX]\lim_{x \to +\infty} f(x)=+\infty[/TEX]
Để PT có nghiệm thì:
[TEX]2m \geq Minf(x)=-1 \Rightarrow m \geq -\frac{1}{2}[/TEX]