Tính đạo hàm của hs sau: a) y= x-1/5x-2 b) y= 2x+3/7+3x c) y= x^2+2x+3/3-4x d) y= x^2+7x+3/x^2-3x

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1. Đáp án:

   $\begin{array}{l}
   a)y = \frac{{x – 1}}{{5x – 2}}\\
    \Rightarrow y’ = \frac{{\left( {x – 1} \right)’.\left( {5x – 2} \right) – \left( {5x – 2} \right)’.\left( {x – 1} \right)}}{{{{\left( {5x – 2} \right)}^2}}}\\
    = \frac{{5x – 2 – 5x + 5}}{{{{\left( {5x – 2} \right)}^2}}}\\
    = \frac{3}{{{{\left( {5x – 2} \right)}^2}}}\\
   b)y = \frac{{2x + 3}}{{7 + 3x}} = \frac{{2x + 3}}{{3x + 7}}\\
    \Rightarrow y’ = \frac{{2.\left( {3x + 7} \right) – 3.\left( {3x + 3} \right)}}{{{{\left( {3x + 7} \right)}^2}}}\\
    = \frac{5}{{{{\left( {3x + 7} \right)}^2}}}\\
   c)y = \frac{{{x^2} + 2x + 3}}{{3 – 4x}}\\
    \Rightarrow y’ = \frac{{\left( {2x + 2} \right).\left( {3 – 4x} \right) – \left( { – 4} \right).\left( {{x^2} + 2x + 3} \right)}}{{{{\left( {3 – 4x} \right)}^2}}}\\
    = \frac{{ – 4{x^2} + 6x + 18}}{{{{\left( {3 – 4x} \right)}^2}}}\\
   d)y = \frac{{{x^2} + 7x + 3}}{{{x^2} – 3x}}\\
    \Rightarrow y’ = \frac{{\left( {2x + 7} \right)\left( {{x^2} – 3x} \right) – \left( {2x – 3} \right).\left( {{x^2} + 7x + 3} \right)}}{{{{\left( {{x^2} – 3x} \right)}^2}}}\\
    = \frac{{ – 8{x^2} – 6x + 9}}{{{{\left( {{x^2} – 3x} \right)}^2}}}
   \end{array}$

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