Đáp án đúng là: D

Ta có: \(\sqrt {1 + 2{\rm{x}}} \cdot \sqrt[3]{{1 + 3{\rm{x}}}} \cdot \sqrt[4]{{1 + 4{\rm{x}}}} - 1\)

\( = \sqrt {1 + 2{\rm{x}}} - \sqrt {1 + 2{\rm{x}}} + \sqrt {1 + 2{\rm{x}}} .\sqrt[3]{{1 + 3{\rm{x}}}} - \sqrt {1 + 2{\rm{x}}} \cdot \sqrt[3]{{1 + 3{\rm{x}}}} + \sqrt {1 + 2{\rm{x}}} \cdot \sqrt[3]{{1 + 3{\rm{x}}}} \cdot \sqrt[4]{{1 + 4{\rm{x}}}} - 1\)\( = \left( {\sqrt {1 + 2{\rm{x}}} - 1} \right) + \sqrt {1 + 2{\rm{x}}} \cdot \left( {\sqrt[3]{{1 + 3{\rm{x}}}} - 1} \right) + \sqrt {1 + 2{\rm{x}}} \cdot \sqrt[3]{{1 + 3{\rm{x}}}}\left( {1 + \sqrt[4]{{1 + 4{\rm{x}}}}} \right)\)

Suy ra: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2{\rm{x}}} .\sqrt[3]{{1 + 3{\rm{x}}}}.\sqrt[4]{{1 + 4{\rm{x}}}} - 1}}{x}\)

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

Ta có: \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt {1 + 2{\rm{x}}} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {1 + 2{\rm{x}}} - 1} \right).\left( {\sqrt {1 + 2{\rm{x}}} + 1} \right)}}{{x\left( {\sqrt {1 + 2{\rm{x}}} + 1} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{\rm{x}}}}{{x\left( {\sqrt {1 + 2{\rm{x}}} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{2}{{\left( {\sqrt {1 + 2{\rm{x}}} + 1} \right)}} = \frac{2}{{1 + 1}} = 1\)

Ta có:

\(\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2{\rm{x}}} .\frac{{\sqrt[3]{{1 + 3{\rm{x}}}} - 1}}{x}} \right)\)

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

\( = \mathop {\lim }\limits_{x \to 0} \left\{ {\sqrt {1 + 2{\rm{x}}} .\frac{{3{\rm{x}}}}{{x\left[ {{{\left( {\sqrt[3]{{1 + 3{\rm{x}}}}} \right)}^2} + \sqrt[3]{{1 + 3{\rm{x}}}} + 1} \right]}}} \right\}\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{3\sqrt {1 + 2{\rm{x}}} }}{{\left[ {{{\left( {\sqrt[3]{{1 + 3{\rm{x}}}}} \right)}^2} + \sqrt[3]{{1 + 3{\rm{x}}}} + 1} \right]}}} \right) = \frac{{3.1}}{{1 + 1 + 1}} = 1\)

Ta có: \(\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2{\rm{x}}} .\sqrt[3]{{1 + 3{\rm{x}}}}.\frac{{\sqrt[4]{{1 + 4{\rm{x}}}} - 1}}{x}} \right)\)

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

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

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{4\sqrt {1 + 2{\rm{x}}} .\sqrt[3]{{1 + 3{\rm{x}}}}}}{{\left[ {{{\left( {\sqrt[4]{{1 + 4{\rm{x}}}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4{\rm{x}}}}} \right)}^2} + \left( {\sqrt[4]{{1 + 4{\rm{x}}}}} \right) + 1} \right]}}} \right) = \frac{{4.1.1}}{{1 + 1 + 1 + 1}} = 1\)

Suy ra \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2{\rm{x}}} .\sqrt[3]{{1 + 3{\rm{x}}}}.\sqrt[4]{{1 + 4{\rm{x}}}} - 1}}{x} = 1 + 1 + 1 = 3\)

Vậy ta chọn đáp án D.

Rút gọn biểu thức P = x^(1/3) căn bậc sáu x, x > 0 A. P = x^2 B. P = căn bậc hai x