a) Ta có: \(M\left( {x;\frac{{2x + 1}}{x}} \right)\); \(H\left( {x;2} \right)\).
Do đó, \(MH = \sqrt {{{\left( {x - x} \right)}^2} + {{\left( {2 - \frac{{2x + 1}}{x}} \right)}^2}} = \sqrt {{{\left( {\frac{{2x - 2x - 1}}{x}} \right)}^2}} = \frac{1}{x}\) (do \(x > 0\))
b) Ta có: \(\mathop {\lim }\limits_{x \to + \infty } \frac{1}{x} = 0\). Do đó, khi \(x \to + \infty \) thì \(MH \to 0\).
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