C113

Ta có: \(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{1}{{a + b + c}} \Longrightarrow \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{{a + b + c}} - \dfrac{1}{c}\)

\(\begin{array}{l} \Longrightarrow \left( {a + b} \right)\left( {a + b + c} \right)c = abc - ab\left( {a + b + c} \right)\\ \Longrightarrow \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) = 0 \end{array}\)

............

Ta có: \(M\left( {0;y} \right)\) 

Lại có: \(\overrightarrow {MA} \left( {1;1 - y} \right),\overrightarrow {MB} \left( {2; - 2 - y} \right)\)

Theo yêu cầu bài toán, suy ra: \({1^2} + {\left( {1 - y} \right)^2} = {2^2} + {\left( {2 + y} \right)^2} \Leftrightarrow 1 + 1 - 2y + {y^2} = 4 + 4 + 4y + {y^2} \Leftrightarrow y = - 1\)

Nên \(M\left( {0; - 1} \right)\)

Vậy \(a = 0,b = - 1 \Rightarrow a + b = 0 + \left( { - 1} \right) = - 1\)

\(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 100} + x} \right) = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {\sqrt {{x^2} + 100} + x} \right)\left( {\sqrt {{x^2} + 100} - x} \right)}}{{\sqrt {{x^2} + 100} - x}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{100}}{{ - x\sqrt {1 + \dfrac{{100}}{x} - x} }} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{{100}}{x}}}{{ - \sqrt {1 + \dfrac{{100}}{x} - 1} }} = 0\)

\(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - x + 1} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} + x + 1 - {x^2}}}{{\sqrt {{x^2} + x + 1} + x}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{1 + \dfrac{1}{x}}}{{\sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}} + 1} }} = \dfrac{1}{2}\)

Phương trình có 2 nghiệm trái dấu khi $ac<0$ hay \(m\left( {m - 4} \right) < 0 \Leftrightarrow 0 < m < 4\)