Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\lim \frac{{3{n^2} - 2n - 1}}{{{n^2} - 1}} = \lim \frac{{3 - \frac{2}{n} - \frac{1}{{{n^2}}}}}{{1 - \frac{1}{{{n^2}}}}} = \frac{{3 - 0 - 0}}{{1 - 0}} = 3\\
b,\\
\lim \frac{{\sqrt {2{n^2} + 3n + 1} }}{{3n - 2}} = \lim \frac{{\sqrt {{n^2}\left( {2 + \frac{3}{n} + \frac{1}{{{n^2}}}} \right)} }}{{3n - 2}} = \lim \frac{{n\sqrt {2 + \frac{3}{n} + \frac{1}{{{n^2}}}} }}{{3n - 2}} = \lim \frac{{\sqrt {2 + \frac{3}{n} + \frac{1}{{{n^2}}}} }}{{3 - \frac{2}{n}}} = \frac{{\sqrt {2 + 0 + 0} }}{{3 - 0}} = \frac{{\sqrt 2 }}{3}\\
c,\\
\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 3x + 2}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x - 2} \right) = 1 - 2 = - 1\\
d,\\
\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {2x - 3} - 1}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {\sqrt {2x - 3} - 1} \right)\left( {\sqrt {2x - 3} + 1} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {2x - 3} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\left( {2x - 3} \right) - {1^2}}}{{\left( {x - 2} \right)\left( {\sqrt {2x - 3} + 1} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{2\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {2x - 3} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{2}{{\sqrt {2x - 3} + 1}} = \frac{2}{{\sqrt {2.2 - 3} + 1}} = 1\\
2,\\
a,\\
y = {\left( {2x - 3} \right)^5} \Rightarrow y' = 5.\left( {2x - 3} \right)'.{\left( {2x - 3} \right)^4} = 5.2.{\left( {2x - 3} \right)^4} = 10.{\left( {2x - 3} \right)^4}\\
b,\\
y = \frac{{3x - 2}}{{x + 1}}\\
\Rightarrow y' = \frac{{\left( {3x - 2} \right)'.\left( {x + 1} \right) - \left( {x + 1} \right)'.\left( {3x - 2} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\
= \frac{{3\left( {x + 1} \right) - 1.\left( {3x - 2} \right)}}{{{{\left( {x + 1} \right)}^2}}} = \frac{5}{{{{\left( {x + 1} \right)}^2}}}\\
b,\\
y = {\sin ^3}\left( {2x - \frac{\pi }{6}} \right)\\
\Rightarrow y' = 3.\left[ {\sin \left( {2x - \frac{\pi }{6}} \right)} \right]'.{\sin ^2}\left( {2x - \frac{\pi }{6}} \right)\\
= 3.\left( {2x - \frac{\pi }{6}} \right)'.\cos \left( {2x - \frac{\pi }{6}} \right).{\sin ^2}\left( {2x - \frac{\pi }{6}} \right)\\
= 6.\cos \left( {2x - \frac{\pi }{6}} \right).{\sin ^2}\left( {2x - \frac{\pi }{6}} \right)\\
d,\\
y = \cos \left( {{x^2} - 3x - 1} \right)\\
\Rightarrow y' = \left( {{x^2} - 3x - 1} \right)'.\left( { - \sin \left( {{x^2} - 3x - 1} \right)} \right)\\
= - \left( {2x - 3} \right)\sin \left( {{x^2} - 3x - 1} \right)
\end{array}\)
