Giải thích các bước giải:
Tìm đa thức
Để tìm được $\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt[3]{{6x - 5}} - \sqrt {4x - 3} }}{{{{\left( {x - 1} \right)}^2}}}$ cần thêm bớt nhị thức bậc nhất $ax+b$ và tử thức để sau khi liên hợp thì $(x-1)^2$ bị triệt tiêu.
Khi đó:
$\left\{ \begin{array}{l}
ax + b = \sqrt {4x - 3} \left( 1 \right)\\
ax + b = \sqrt[3]{{6x - 5}}
\end{array} \right.$
Xét $(1)$ ta có:
Khi $x=1$ thì $a+b=1(*)$
Lại có:
$\begin{array}{l}
\left( 1 \right) \Leftrightarrow {\left( {ax + b} \right)^2} = 4x - 3\\
\Leftrightarrow {a^2}{x^2} + 2\left( {ab - 2} \right)x + {b^2} + 3 = 0\left( 2 \right)
\end{array}$
Để $(2)$ có nghiệm kép
$\begin{array}{l}
\Leftrightarrow \Delta ' = 0\\
\Leftrightarrow {\left( {ab - 2} \right)^2} - {a^2}\left( {{b^2} + 3} \right) = 0\\
\Leftrightarrow 3{a^2} + 4ab - 4 = 0\left( {**} \right)
\end{array}$
Từ $\left( * \right),\left( {**} \right) \Rightarrow \left\{ \begin{array}{l}
a + b = 1\\
3{a^2} + 4ab - 4 = 0
\end{array} \right.$
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
a + b = 1\\
3{a^2} + 4a\left( {1 - a} \right) - 4 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a + b = 1\\
- {a^2} + 4a - 4 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a + b = 1\\
{\left( {a - 2} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 2\\
b = - 1
\end{array} \right.
\end{array}$
Như vậy: Đa thức cần thêm bớt là: $2x-1$
Tìm giới hạn
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt[3]{{6x - 5}} - \sqrt {4x - 3} }}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt[3]{{6x - 5}} - \left( {2x - 1} \right) + 2x - 1 - \sqrt {4x - 3} }}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {6x - 5} \right) - {{\left( {2x - 1} \right)}^3}}}{{\left( {{{\left( {\sqrt[3]{{6x - 5}}} \right)}^2} + \sqrt[3]{{6x - 5}}\left( {2x - 1} \right) + {{\left( {2x - 1} \right)}^2}} \right){{\left( {x - 1} \right)}^2}}} + \mathop {\lim }\limits_{x \to 1} \dfrac{{{{\left( {2x - 1} \right)}^2} - \left( {4x - 3} \right)}}{{\left( {2x - 1 + \sqrt {4x - 3} } \right){{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 4\left( {2{x^3} - 3{x^2} + 1} \right)}}{{\left( {{{\left( {\sqrt[3]{{6x - 5}}} \right)}^2} + \sqrt[3]{{6x - 5}}\left( {2x - 1} \right) + {{\left( {2x - 1} \right)}^2}} \right){{\left( {x - 1} \right)}^2}}} + \mathop {\lim }\limits_{x \to 1} \dfrac{{4\left( {{x^2} - 2x + 1} \right)}}{{\left( {2x - 1 + \sqrt {4x - 3} } \right){{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 4\left( {2x + 1} \right)}}{{{{\left( {\sqrt[3]{{6x - 5}}} \right)}^2} + \sqrt[3]{{6x - 5}}\left( {2x - 1} \right) + {{\left( {2x - 1} \right)}^2}}} + \mathop {\lim }\limits_{x \to 1} \dfrac{4}{{2x - 1 + \sqrt {4x - 3} }}\\
= \dfrac{{ - 4\left( {2.1 + 1} \right)}}{{{{\left( {\sqrt[3]{{6.1 - 5}}} \right)}^2} + \sqrt[3]{{6.1 - 5}}\left( {2.1 - 1} \right) + {{\left( {2.1 - 1} \right)}^2}}} + \dfrac{4}{{2.1 - 1 + \sqrt {4.1 - 3} }}\\
= \dfrac{{ - 12}}{3} + \dfrac{4}{2}\\
= - 2
\end{array}$
