A. \(y = \dfrac{{3x + 10}}{{5x + 7}}\)
B. \(y = \dfrac{{ - x + 1}}{{5x - 3}}\)
C. \(y = \dfrac{{ - x - 8}}{{x + 3}}\)
D. \(y = \dfrac{{3x + 5}}{{x + 1}}\)
C
Đáp án A:
\(\begin{array}{l}y' = \dfrac{{3\left( {5x + 7} \right) - 5\left( {3x + 10} \right)}}{{{{\left( {5x + 7} \right)}^2}}}\\ = - \dfrac{{29}}{{{{\left( {5x + 7} \right)}^2}}} < 0\left( L \right)\end{array}\)
Đáp án B:
\(\begin{array}{l}y' = \dfrac{{ - 1\left( {5x - 3} \right) - 5\left( { - x + 1} \right)}}{{{{\left( {5x - 3} \right)}^2}}}\\ = - \dfrac{2}{{{{\left( {5x - 3} \right)}^2}}} < 0\left( L \right)\end{array}\)
Đáp án C:
\(\begin{array}{l}y' = \dfrac{{ - 1\left( {x + 3} \right) - \left( { - x - 8} \right)}}{{{{\left( {x + 3} \right)}^2}}}\\ = \dfrac{5}{{{{\left( {x + 3} \right)}^2}}} > 0\left( {TM} \right)\end{array}\)
Đáp án D:
\(\begin{array}{l}y' = \dfrac{{3\left( {x + 1} \right) - \left( {3x + 5} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\ = - \dfrac{2}{{{{\left( {x + 1} \right)}^2}}} < 0\left( L \right)\end{array}\)
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