Cho (E): \(\frac{{{x^2}}}{{...

B

- Hướng dẫn giải

Vì \(M\left( {x;\,y} \right) \in \left( E \right)\) nên \(\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1\) và \(OM = \sqrt {{x^2} + {y^2}} \).

Ta có \(\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{16}} \le \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} \le \frac{{{x^2}}}{9} + \frac{{{y^2}}}{9}\)

\(\begin{array}{l} \Leftrightarrow \frac{{O{M^2}}}{{16}} \le 1 \le \frac{{O{M^2}}}{9}\\ \Leftrightarrow 9 \le O{M^2} \le 16\\ \Leftrightarrow 3 \le OM \le 4\end{array}\)