mathop lim limitsx to 1 căn [3]3x - 2 - căn 4x^2 - x - 2 x^2 - 3x + 2
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Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{3x - 2}} - \sqrt {4{x^2} - x - 2} }}{{{x^2} - 3x + 2}}\)
Đáp án đúng: D
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{3x - 2}} - \sqrt {4{x^2} - x - 2} }}{{{x^2} - 3x + 2}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{3x - 2}} - 1 + 1 - \sqrt {4{x^2} - x - 2} }}{{{x^2} - 3x + 2}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{3x - 2}} - 1}}{{{x^2} - 3x + 2}} + \mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt {4{x^2} - x - 2} }}{{{x^2} - 3x + 2}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt[3]{{3x - 2}} - 1} \right)\left( {{{\sqrt[3]{{3x - 2}}}^2} + \sqrt[3]{{3x - 2}} + 1} \right)}}{{\left( {{x^2} - 3x + 2} \right)\left( {{{\sqrt[3]{{3x - 2}}}^2} + \sqrt[3]{{3x - 2}} + 1} \right)}} + \mathop {\lim }\limits_{x \to 1} \frac{{\left( {1 - \sqrt {4{x^2} - x - 2} } \right)\left( {1 + \sqrt {4{x^2} - x - 2} } \right)}}{{\left( {{x^2} - 3x + 2} \right)\left( {1 + \sqrt {4{x^2} - x - 2} } \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{3\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {{{\sqrt[3]{{3x - 2}}}^2} + \sqrt[3]{{3x - 2}} + 1} \right)}} + \mathop {\lim }\limits_{x \to 1} \frac{{ - 4{x^2} + x + 3}}{{\left( {{x^2} - 3x + 2} \right)\left( {1 + \sqrt {4{x^2} - x - 2} } \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{3}{{\left( {x - 2} \right)\left( {{{\sqrt[3]{{3x - 2}}}^2} + \sqrt[3]{{3x - 2}} + 1} \right)}} - \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {4x + 3} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {1 + \sqrt {4{x^2} - x - 2} } \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{3}{{\left( {x - 2} \right)\left( {{{\sqrt[3]{{3x - 2}}}^2} + \sqrt[3]{{3x - 2}} + 1} \right)}} - \mathop {\lim }\limits_{x \to 1} \frac{{4x + 3}}{{\left( {x - 2} \right)\left( {1 + \sqrt {4{x^2} - x - 2} } \right)}}\\ = \frac{3}{{ - 1.\left( {1 + 1 + 1} \right)}} - \frac{7}{{ - 1.\left( {1 + 1} \right)}} = \frac{5}{2}\end{array}\)
Chọn D.
