Đáp án đúng: A
Giải chi tiết:\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{\left( {1 + ax} \right)\left( {1 + bx} \right)\left( {1 + cx} \right)}} - 1}}{x}\,\,\,\,\left( {abc \ne 0} \right)\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}}\sqrt[n]{{1 + bx}}\sqrt[n]{{1 + cx}} - 1}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[n]{{1 + ax}} - 1} \right)\sqrt[n]{{1 + bx}}\sqrt[n]{{1 + cx}}}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[n]{{1 + bx}} - 1} \right)\sqrt[n]{{1 + cx}}}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + cx}} - 1}}{x}\\ = {L_1} + {L_2} + {L_3}\end{array}\)
\(\begin{array}{l}{L_1} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[n]{{1 + ax}} - 1} \right)\sqrt[n]{{1 + bx}}\sqrt[n]{{1 + cx}}}}{x}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[n]{{1 + ax}} - 1} \right)\sqrt[n]{{1 + bx}}\sqrt[n]{{1 + cx}}\left( {{{\sqrt[n]{{1 + ax}}}^{n - 1}} + {{\sqrt[n]{{1 + ax}}}^{n - 2}} + ... + \sqrt[n]{{1 + ax}} + 1} \right)}}{{x\left( {{{\sqrt[n]{{1 + ax}}}^{n - 1}} + {{\sqrt[n]{{1 + ax}}}^{n - 2}} + ... + \sqrt[n]{{1 + ax}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{ax.\sqrt[n]{{1 + bx}}\sqrt[n]{{1 + cx}}}}{{x\left( {{{\sqrt[n]{{1 + ax}}}^{n - 1}} + {{\sqrt[n]{{1 + ax}}}^{n - 2}} + ... + \sqrt[n]{{1 + ax}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{a\sqrt[n]{{1 + bx}}\sqrt[n]{{1 + cx}}}}{{{{\sqrt[n]{{1 + ax}}}^{n - 1}} + {{\sqrt[n]{{1 + ax}}}^{n - 2}} + ... + \sqrt[n]{{1 + ax}} + 1}}\\\,\,\,\,\,\, = \frac{{a.1.1}}{{1 + 1 + 1 + ... + 1\,\,\,\left( {n\,\,chu\,\,so\,\,1} \right)}} = \frac{a}{n}\end{array}\)
\(\begin{array}{l}{L_2} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[n]{{1 + bx}} - 1} \right)\sqrt[n]{{1 + cx}}}}{x}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[n]{{1 + bx}} - 1} \right)\sqrt[n]{{1 + cx}}\left( {{{\sqrt[n]{{1 + bx}}}^{n - 1}} + {{\sqrt[n]{{1 + bx}}}^{n - 2}} + ... + \sqrt[n]{{1 + bx}} + 1} \right)}}{{x\left( {{{\sqrt[n]{{1 + bx}}}^{n - 1}} + {{\sqrt[n]{{1 + bx}}}^{n - 2}} + ... + \sqrt[n]{{1 + bx}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{bx\sqrt[n]{{1 + cx}}}}{{x\left( {{{\sqrt[n]{{1 + bx}}}^{n - 1}} + {{\sqrt[n]{{1 + bx}}}^{n - 2}} + ... + \sqrt[n]{{1 + bx}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{b\sqrt[n]{{1 + cx}}}}{{{{\sqrt[n]{{1 + bx}}}^{n - 1}} + {{\sqrt[n]{{1 + bx}}}^{n - 2}} + ... + \sqrt[n]{{1 + bx}} + 1}}\\\,\,\,\,\, = \frac{{b.1}}{{1 + 1 + 1 + ... + 1\,\,\,\left( {n\,\,chu\,\,so\,\,1} \right)}} = \frac{b}{n}\end{array}\) 
\(\begin{array}{l}{L_3} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + cx}} - 1}}{x}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[n]{{1 + cx}} - 1} \right)\left( {{{\sqrt[n]{{1 + cx}}}^{n - 1}} + {{\sqrt[n]{{1 + cx}}}^{n - 2}} + ... + \sqrt[n]{{1 + cx}} + 1} \right)}}{{x\left( {{{\sqrt[n]{{1 + cx}}}^{n - 1}} + {{\sqrt[n]{{1 + cx}}}^{n - 2}} + ... + \sqrt[n]{{1 + cx}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{cx}}{{x\left( {{{\sqrt[n]{{1 + cx}}}^{n - 1}} + {{\sqrt[n]{{1 + cx}}}^{n - 2}} + ... + \sqrt[n]{{1 + cx}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{c}{{{{\sqrt[n]{{1 + cx}}}^{n - 1}} + {{\sqrt[n]{{1 + cx}}}^{n - 2}} + ... + \sqrt[n]{{1 + cx}} + 1}}\\\,\,\,\,\,\, = \frac{c}{{1 + 1 + 1 + ... + 1\,\,\left( {n\,\,chu\,\,so\,\,1} \right)}} = \frac{c}{n}\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{\left( {1 + ax} \right)\left( {1 + bx} \right)\left( {1 + cx} \right)}} - 1}}{x} = \frac{{a + b + c}}{n}\).
Chọn A.