Do \(x \ge - 1 \Rightarrow x + 1 \ge 0 \Rightarrow y = \frac{{x + 1}}{{\sqrt {{x^2} + 1} }} \ge 0 = y\left( { - 1} \right)\)
Vậy \(\mathop {\min y}\limits_{\left[ { - 1; + \infty } \right)} = y\left( { - 1} \right) = 0\)
Do \(x \ge - 1 \Rightarrow x + 1 \ge 0 \Rightarrow y = \frac{{x + 1}}{{\sqrt {{x^2} + 1} }} = \sqrt {\frac{{{{\left( {x + 1} \right)}^2}}}{{{x^2} + 1}}} \)
\(\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{{{x^2} + 2x + 1}}{{{x^2} + 1}}} = \sqrt {2 + \frac{{{x^2} + 2x + 1}}{{{x^2} + 1}} - 2} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {2 + \left( {\frac{{{x^2} + 2x + 1}}{{{x^2} + 1}} - 2} \right)} = \sqrt {2 - \frac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + 1}}} \le \sqrt 2 = y\left( 1 \right)
\end{array}\)
Vậy \(\mathop {\max y}\limits_{\left[ { - 1; + \infty } \right)} = y\left( 1 \right) = \sqrt 2 \)
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