Biết \(\mathop {\lim }\limits_{x \to 3} \dfrac{{x + 1 - \sqrt {5x + 1} }}{{x - \sqrt {4x - 3} }} = \dfrac{a}{b}\).
Biết \(\mathop {\lim }\limits_{x \to 3} \dfrac{{x + 1 - \sqrt {5x + 1} }}{{x - \sqrt {4x - 3} }} = \dfrac{a}{b}\). Tính a + b.
Đáp án đúng là:
Quảng cáo
Ta có \(\mathop {\lim }\limits_{x \to 3} \dfrac{{x + 1 - \sqrt {5x + 1} }}{{x - \sqrt {4x - 3} }}\)
\( = \mathop {\lim }\limits_{x \to 3} \dfrac{{\left( {{{(x + 1)}^2} - 5x - 1} \right)(x + \sqrt {4x - 3} )}}{{\left( {{x^2} - 4x + 3} \right)(x + 1 + \sqrt {5x + 1} )}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 3} \dfrac{{\left( {{x^2} - 3x} \right)(x + \sqrt {4x - 3} )}}{{\left( {{x^2} - 4x + 3} \right)(x + 1 + \sqrt {5x + 1} )}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{x(x + \sqrt {4x - 3} )}}{{(x - 1)(x + 1 + \sqrt {5x + 1} )}} = \dfrac{{18}}{{16}} = \dfrac{9}{8}\end{array}\)
\( \Rightarrow a + b = 17\).
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