đk: -1 ≤x≤1
áp dụng bđt bunhiacopxki có:
\(\left(\sqrt{1-x}+2\sqrt{x+1}\right)^2\le\left(1^2+2^2\right)\left(1-x+x+1\right)=10\)
\(\Rightarrow Q^2\le10\Rightarrow-\sqrt{10}\le Q\le\sqrt{10}\)
dấu ''='' xayr ra khi: \(\sqrt{1-x}=\frac{\sqrt{x+1}}{2}\Leftrightarrow2\sqrt{1-x}=\sqrt{x+1}\Leftrightarrow x=\frac{3}{5}\)