Phương pháp giải:

Cho hàm số \(y = f\left( x \right)\).

+) Nếu \(\mathop {\lim }\limits_{x \to \infty } y = {y_0} \Rightarrow y = {y_0}\) là TCN của đồ thị hàm số.

+) Nếu \(\mathop {\lim }\limits_{x \to {x_0}} y = \infty {\rm{\;}} \Rightarrow x = {x_0}\) là TCĐ của đồ thị hàm số.

Lời giải chi tiết:

TXĐ: \(x \ge {\rm{\;}} - \dfrac{1}{3};{\mkern 1mu} {\mkern 1mu} x \ne 1;{\mkern 1mu} {\mkern 1mu} x \ne 2\). Ta có:

\(\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to + \infty } y = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x + 1 - \sqrt {3x + 1} }}{{{x^2} - 3x + 2}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{1}{x} + \dfrac{1}{{{x^2}}} - \sqrt {\dfrac{3}{{{x^3}}} + \dfrac{1}{{{x^4}}}} }}{{1 - \dfrac{3}{x} + \dfrac{2}{{{x^2}}}}} = 0}\\{\mathop {\lim }\limits_{x \to - \infty } y = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x + 1 - \sqrt {3x + 1} }}{{{x^2} - 3x + 2}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{1}{x} + \dfrac{1}{{{x^2}}} - \sqrt {\dfrac{3}{{{x^3}}} + \dfrac{1}{{{x^4}}}} }}{{1 - \dfrac{3}{x} + \dfrac{2}{{{x^2}}}}} = 0}\end{array}\)

Do đó đồ thị hàm số có TCN \(y = 0\).

\(\begin{array}{*{20}{l}}\begin{array}{l}y = \dfrac{{x + 1 - \sqrt {3x + 1} }}{{{x^2} - 3x + 2}}\\ = \dfrac{{\left( {x + 1 - \sqrt {3x + 1} } \right)\left( {x + 1 + \sqrt {3x + 1} } \right)}}{{\left( {x + 1 + \sqrt {3x + 1} } \right)\left( {{x^2} - 3x + 2} \right)}}\\ = \dfrac{{{{\left( {x + 1} \right)}^2} - \left( {3x + 1} \right)}}{{\left( {x + 1 + \sqrt {3x + 1} } \right)\left( {{x^2} - 3x + 2} \right)}}\end{array}\\\begin{array}{l} = \dfrac{{{x^2} - x}}{{\left( {x + 1 + \sqrt {3x + 1} } \right)\left( {{x^2} - 3x + 2} \right)}}\\ = \dfrac{{x\left( {x - 1} \right)}}{{\left( {x + 1 + \sqrt {3x + 1} } \right)\left( {x - 1} \right)\left( {x - 2} \right)}}\\ = \dfrac{x}{{\left( {x + 1 + \sqrt {3x + 1} } \right)\left( {x - 2} \right)}}\end{array}\end{array}\)

Ta có

\(\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to {2^ + }} y = \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{x}{{\left( {x + 1 + \sqrt {3x + 1} } \right)\left( {x - 2} \right)}} = + \infty }\\{\mathop {\lim }\limits_{x \to {2^ - }} y = \mathop {\lim }\limits_{x \to {2^ - }} \dfrac{x}{{\left( {x + 1 + \sqrt {3x + 1} } \right)\left( {x - 2} \right)}} = - \infty }\end{array}\), do đó đồ thị hàm số có TCĐ .

Xét phương trình

\(\begin{array}{*{20}{l}}{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} x + 1 + \sqrt {3x + 1} = 0 \Leftrightarrow \sqrt {3x + 1} = - x - 1 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{ - x - 1 \ge 0}\\{3x + 1 = {{\left( { - x - 1} \right)}^2}}\end{array}} \right.}\\{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{x \le - 1}\\{3x + 1 = {x^2} + 2x + 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{x \le - 1}\\{{x^2} - x = 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{x \le - 1}\\{\left[ {\begin{array}{*{20}{l}}{x = 0}\\{x = 1}\end{array}} \right.}\end{array}} \right. \Leftrightarrow x \in \emptyset }\end{array}\)

Vậy hàm số có 1 TCN \(y = 0\) và 1 TCĐ \(x = 2\).

Chọn A.