Đáp án:

\[B = \mathop {\lim }\limits_{x \to  - 1} \frac{{\sqrt {5 + 4x}  - \sqrt[3]{{7 + 6x}}}}{{{x^3} + {x^2} - x - 1}} =  - 1\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
B = \mathop {\lim }\limits_{x \to  - 1} \frac{{\sqrt {5 + 4x}  - \sqrt[3]{{7 + 6x}}}}{{{x^3} + {x^2} - x - 1}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {\sqrt {5 + 4x}  - \left( {2x + 3} \right)} \right) + \left( {\left( {2x + 3} \right) - \sqrt[3]{{7 + 6x}}} \right)}}{{{x^2}\left( {x + 1} \right) - \left( {x + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{\frac{{5 + 4x - {{\left( {2x + 3} \right)}^2}}}{{\sqrt {5 + 4x}  + 2x + 3}} + \frac{{{{\left( {2x + 3} \right)}^3} - \left( {7 + 6x} \right)}}{{{{\left( {2x + 3} \right)}^2} + \left( {2x + 3} \right).\sqrt[3]{{7 + 6x}} + {{\sqrt[3]{{7 + 6x}}}^2}}}}}{{\left( {x + 1} \right)\left( {{x^2} - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{\frac{{5 + 4x - 4{x^2} - 12x - 9}}{{\sqrt {5 + 4x}  + 2x + 3}} + \frac{{8{x^3} + 36{x^2} + 54x + 27 - 7 - 6x}}{{{{\left( {2x + 3} \right)}^2} + \left( {2x + 3} \right).\sqrt[3]{{7 + 6x}} + {{\sqrt[3]{{7 + 6x}}}^2}}}}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{\frac{{ - 4{{\left( {x + 1} \right)}^2}}}{{\sqrt {5 + 4x}  + 2x + 3}} + \frac{{4{{\left( {x + 1} \right)}^2}\left( {2x + 5} \right)}}{{{{\left( {2x + 3} \right)}^2} + \left( {2x + 3} \right).\sqrt[3]{{7 + 6x}} + {{\sqrt[3]{{7 + 6x}}}^2}}}}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \left[ {\frac{1}{{x - 1}}.\left( {\frac{{ - 4}}{{\sqrt {5 + 4x}  + 2x + 3}} + \frac{{4\left( {2x + 5} \right)}}{{{{\left( {2x + 3} \right)}^2} + \left( {2x + 3} \right).\sqrt[3]{{7 + 6x}} + {{\sqrt[3]{{7 + 6x}}}^2}}}} \right)} \right]\\
 = \frac{1}{{ - 1 - 1}}.\left( {\frac{{ - 4}}{{1 + 1}} + \frac{{4.3}}{{1 + 1 + 1}}} \right)\\
 = \frac{{ - 1}}{2}.\left( { - 2 + 4} \right)\\
 =  - 1
\end{array}\)