Đáp án:
\[{L_9} = \frac{{ - 1}}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{L_9} = \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\left( {\sqrt {{x^2} + 2x} + x - 2\sqrt {{x^2} + x} } \right)} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\left( {\left( {\sqrt {{x^2} + 2x} - \sqrt {{x^2} + x} } \right) + \left( {x - \sqrt {{x^2} + x} } \right)} \right)} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\left( {\dfrac{{\left( {{x^2} + 2x} \right) - \left( {{x^2} + x} \right)}}{{\sqrt {{x^2} + 2x} + \sqrt {{x^2} + x} }} + \dfrac{{{x^2} - \left( {{x^2} + x} \right)}}{{x + \sqrt {{x^2} + x} }}} \right)} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\left( {\dfrac{x}{{\sqrt {{x^2} + 2x} + \sqrt {{x^2} + x} }} - \dfrac{x}{{x + \sqrt {{x^2} + x} }}} \right)} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {{x^2}.\left( {\dfrac{1}{{\sqrt {{x^2} + 2x} + \sqrt {{x^2} + x} }} - \dfrac{1}{{x + \sqrt {{x^2} + x} }}} \right)} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {{x^2}.\dfrac{{x - \sqrt {{x^2} + 2x} }}{{\left( {\sqrt {{x^2} + 2x} + \sqrt {{x^2} + x} } \right).\left( {x + \sqrt {{x^2} + x} } \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {{x^2}.\dfrac{{{x^2} - \left( {{x^2} + 2x} \right)}}{{\left( {x + \sqrt {{x^2} + 2x} } \right).\left( {\sqrt {{x^2} + 2x} + \sqrt {{x^2} + x} } \right).\left( {x + \sqrt {{x^2} + x} } \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {{x^2}.\dfrac{{ - 2x}}{{\left( {x + \sqrt {{x^2} + 2x} } \right).\left( {\sqrt {{x^2} + 2x} + \sqrt {{x^2} + x} } \right).\left( {x + \sqrt {{x^2} + x} } \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{ - 2{x^3}}}{{\left( {x + \sqrt {{x^2} + 2x} } \right).\left( {\sqrt {{x^2} + 2x} + \sqrt {{x^2} + x} } \right).\left( {x + \sqrt {{x^2} + x} } \right)}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{ - 2}}{{\left( {1 + \sqrt {1 + \frac{2}{x}} } \right)\left( {\sqrt {1 + \frac{2}{x}} + \sqrt {1 + \frac{1}{x}} } \right).\left( {1 + \sqrt {1 + \frac{1}{x}} } \right)}}\\
= \dfrac{{ - 2}}{{\left( {1 + \sqrt 1 } \right).\left( {\sqrt 1 + \sqrt 1 } \right).\left( {1 + \sqrt 1 } \right)}}\\
= \dfrac{{ - 1}}{4}
\end{array}\)
