Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
2,\\
\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {x + 3}  - 2}}{{2{x^2} - 3x + 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {\sqrt {x + 3}  - 2} \right)\left( {\sqrt {x + 3}  + 2} \right)}}{{\left( {2{x^2} - 3x + 1} \right)\left( {\sqrt {x + 3}  + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {x + 3} \right) - {2^2}}}{{\left( {x - 1} \right)\left( {2x - 1} \right)\left( {\sqrt {x + 3}  + 2} \right)}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{x - 1}}{{\left( {x - 1} \right)\left( {2x - 1} \right)\left( {\sqrt {x + 3}  + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to {1^ - }} \frac{1}{{\left( {2x - 1} \right)\left( {\sqrt {x + 3}  + 2} \right)}} = \frac{1}{{\left( {2.1 - 1} \right)\left( {\sqrt {1 + 3}  + 2} \right)}} = \frac{1}{4}\\
\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) =  - 2a + 7
\end{array}\)

Hàm số đã cho liên tục tại \(x = 1\) khi và chỉ khi:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \Leftrightarrow  - 2a + 7 = \frac{1}{4} \Leftrightarrow a = \frac{{27}}{8}\\
3,\\
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3x + 1}  - \sqrt {x + 3} }}{{{x^2} - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {3x + 1}  - \sqrt {x + 3} } \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}{{\left( {{x^2} - 1} \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {3x + 1} \right) - \left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{2\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{2}{{\left( {x + 1} \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}\\
 = \frac{2}{{\left( {1 + 1} \right)\left( {\sqrt {3.1 + 1}  + \sqrt {1 + 3} } \right)}} = \frac{1}{4}\\
f\left( 1 \right) = 1 - \frac{3}{4} + a = a + \frac{1}{4}\\
 \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right) \Leftrightarrow a + \frac{1}{4} = \frac{1}{4} \Leftrightarrow a = 0\\
4,\\
\mathop {\lim }\limits_{x \to 1} \frac{{x + 1 - \sqrt {x + 3} }}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{x - 1}}{{x - 1}} + \frac{{2 - \sqrt {x + 3} }}{{x - 1}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {1 + \frac{{\left( {2 - \sqrt {x + 3} } \right)\left( {2 + \sqrt {x + 3} } \right)}}{{\left( {x - 1} \right)\left( {2 + \sqrt {x + 3} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {1 + \frac{{{2^2} - \left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {2 + \sqrt {x + 3} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {1 + \frac{{1 - x}}{{\left( {x - 1} \right)\left( {2 + \sqrt {x + 3} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {1 + \frac{{ - 1}}{{2 + \sqrt {x + 3} }}} \right]\\
 = 1 - \frac{1}{{2 + \sqrt {1 + 3} }} = \frac{3}{4}\\
f\left( 1 \right) = \frac{{m.1}}{{1 + {1^2}}} = \frac{m}{2}\\
 \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right) \Leftrightarrow \frac{m}{2} = \frac{3}{4} \Leftrightarrow m = \frac{3}{2}\\
5,\\
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{\sqrt {3x + 1}  - \sqrt {x + 3} }}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {{x^2} - 1} \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}{{\left( {\sqrt {3x + 1}  - \sqrt {x + 3} } \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}{{\left( {3x + 1} \right) - \left( {x + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}{{2\left( {x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 1} \right)\left( {\sqrt {3x + 1}  + \sqrt {x + 3} } \right)}}{2}\\
 = \frac{{\left( {1 + 1} \right)\left( {\sqrt {3.1 + 1}  + \sqrt {1 + 3} } \right)}}{2} = 4\\
f\left( 1 \right) = 5.1 - 1 = 4\\
 \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)
\end{array}\)

Suy ra hàm số đã cho liên tục tại \(x = 1\)