Question

giải và biên luận BPT sau theo tham số m

\(\sqrt{x+2\sqrt{mx-m^2}}+\sqrt{x-2\sqrt{mx-m^2}}\le2\) với m > 0

Answer 1

\(x\ge m\)

\(\sqrt{x-m+2\sqrt{m\left(x-m\right)}+m}+\sqrt{x-m-2\sqrt{m\left(x-m\right)}+m}\le2\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-m}+\sqrt{m}\right)^2}+\sqrt{\left(\sqrt{x-m}-\sqrt{m}\right)^2}\le2\)

\(\Leftrightarrow\sqrt{x-m}+\sqrt{m}+\left|\sqrt{x-m}-\sqrt{m}\right|\le2\)

- Nếu \(\sqrt{x-m}\ge\sqrt{m}\Leftrightarrow x\ge2m\) BPT trở thành:

\(2\sqrt{x-m}\le2\Leftrightarrow x\le m+1\Rightarrow2m\le x\le m+1\)

\(\Rightarrow m+1\ge2m\Rightarrow m\le1\)

- Nếu \(\sqrt{x-m}< \sqrt{m}\Leftrightarrow m\le x< 2m\) BPT trở thành:

\(2\sqrt{m}\le2\Rightarrow m\le1\)

Vậy nếu \(0< m\le1\) thì BPT có nghiệm \(m\le x\le m+1\)