a)  Tính giới hạn: \(\mathop {lim}\limits_{x \to 0} \frac{{({x^2} + 2019)\sqrt[3]{{1 - 2x}} - 2019\sqrt {4x + 1} }}{x}\)b) Viết phương

a) Ta có: 

\(\begin{array}{*{20}{l}}
{L = \mathop {{\rm{lim}}}\limits_{x \to 0} \left( {x\sqrt[3]{{1 - 2x}} + 2019\frac{{\sqrt[3]{{1 - 2x}} - 1}}{x} - 2019\frac{{\sqrt {4x - 1}  - 1}}{x}} \right)}
\end{array}\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} x\sqrt[3]{{1 - 2x}} = 0\)

\(\begin{array}{l}
\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\sqrt[3]{{1 - 2x}} - 1}}{x} = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{ - 2x}}{{x\left( {\sqrt[3]{{{{\left( {1 - 2x} \right)}^2}}} + \sqrt[3]{{1 - 2x}} + 1} \right)}} = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{ - 2}}{{\left( {\sqrt[3]{{{{\left( {1 - 2x} \right)}^2}}} + \sqrt[3]{{1 - 2x}} + 1} \right)}} =  - \frac{2}{3}\\
\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\sqrt {4x + 1}  - 1}}{x} = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{4x}}{{x\left( {\sqrt {4x + 1}  + 1} \right)}} = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{4}{{\sqrt {4x + 1}  + 1}} = 2
\end{array}\)

Vậy \(L = 0 + 2019.\frac{{ - 2}}{3} - 2019.2 =  - 5384\)

b) \(x_0=2\) nên \(y_0=3\)

\(y'=6x^2-4\Rightarrow y'(2)=20\)

Vậy phương trình tiếp tuyến cần tìm là: \(y=20x-37\)