Đáp án:

20) \(\dfrac{9}{8}\)

Giải thích các bước giải:

\(\begin{array}{l}
15)\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x + 7 - 9} \right)\left( {2 + \sqrt {x + 3} } \right)}}{{\left( {4 - x - 3} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 2} \right)\left( {2 + \sqrt {x + 3} } \right)}}{{\left( {1 - x} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 2\left( {2 + \sqrt {x + 3} } \right)}}{{\sqrt {2x + 7}  + 3}}\\
 = \dfrac{{ - 2\left( {2 + \sqrt {1 + 3} } \right)}}{{\sqrt {2.1 + 7}  + 3}} =  - \dfrac{4}{3}\\
17)\mathop {\lim }\limits_{x \to  - 1} \dfrac{{3 + 2x - x - 2}}{{3\left( {x + 1} \right)\left( {\sqrt {3 + 2x}  + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{1 - x}}{{3\left( {x + 1} \right)\left( {\sqrt {3 + 2x}  + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{ - 1}}{{3\left( {\sqrt {3 + 2x}  + \sqrt {x + 2} } \right)}}\\
 = \dfrac{{ - 1}}{{3\left( {\sqrt {3 + 2\left( { - 1} \right)}  + \sqrt { - 1 + 2} } \right)}} =  - \dfrac{1}{6}\\
18)\mathop {\lim }\limits_{x \to 1} \dfrac{{2x + 7 - {x^2} + 8x - 16}}{{\left( {x - 1} \right)\left( {x - 3} \right)\left( {\sqrt {2x + 7}  - x + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - {x^2} + 10x - 9}}{{\left( {x - 1} \right)\left( {x - 3} \right)\left( {\sqrt {2x + 7}  - x + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {9 - x} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x - 3} \right)\left( {\sqrt {2x + 7}  - x + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{9 - x}}{{\left( {x - 3} \right)\left( {\sqrt {2x + 7}  - x + 4} \right)}}\\
 = \dfrac{{9 - 1}}{{\left( {1 - 3} \right)\left( {\sqrt {2.1 + 7}  - 1 + 4} \right)}} =  - \dfrac{2}{3}\\
20)\mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {{x^2} - x - 2} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{\left( {4x + 1 - 9} \right)\left( {x + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x + 1} \right)\left( {x - 2} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{4\left( {x - 2} \right)\left( {x + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x + 1} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{4\left( {x + \sqrt {x + 2} } \right)}}\\
 = \dfrac{{\left( {2 + 1} \right)\left( {\sqrt {4.2 + 1}  + 3} \right)}}{{4\left( {2 + \sqrt {2 + 2} } \right)}} = \dfrac{9}{8}
\end{array}\)