Đáp án đúng:
Giải chi tiết:a) \(y = \dfrac{{2 + \sqrt x }}{{2\sqrt x + 1}}\)
\(\begin{array}{l} \Rightarrow y' = \dfrac{{\dfrac{1}{{2\sqrt x }}\left( {2\sqrt x + 1} \right) - \left( {2 + \sqrt x } \right)\dfrac{1}{{\sqrt x }}}}{{{{\left( {2\sqrt x + 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2\sqrt x + 1 - 4 - 2\sqrt x }}{{2\sqrt x {{\left( {2\sqrt x + 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 3}}{{2\sqrt x {{\left( {2\sqrt x + 1} \right)}^2}}}\end{array}\)
b) \(y = \dfrac{{{x^5} - 3{x^3} + 2x - 1}}{x} = {x^4} - 3{x^2} + 2 - \dfrac{1}{x}\)
\( \Rightarrow y' = 4{x^3} - 6x + \dfrac{1}{{{x^2}}}\)
c) \(y = \dfrac{{{x^2} + 4}}{{2x + 1}}\)
\(\begin{array}{l} \Rightarrow y' = \dfrac{{2x\left( {2x + 1} \right) - \left( {{x^2} + 4} \right).2}}{{{{\left( {2x + 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{4{x^2} + 2x - 2{x^2} - 8}}{{{{\left( {2x + 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2{x^2} + 2x - 8}}{{{{\left( {2x + 1} \right)}^2}}}\end{array}\)
d) \(y = \dfrac{{x + 1}}{{{x^2} - x + 1}}\)
\(\begin{array}{l} \Rightarrow y' = \dfrac{{{x^2} - x + 1 - \left( {x + 1} \right)\left( {2x - 1} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{{x^2} - x + 1 - 2{x^2} - x + 1}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - {x^2} - 2x+2}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\end{array}\)
e) \(y = \dfrac{{{x^2} + 7x + 3}}{{{x^2} - 3x}}\)
\(\begin{array}{l} \Rightarrow y' = \dfrac{{\left( {2x + 7} \right)\left( {{x^2} - 3x} \right) - \left( {{x^2} + 7x + 3} \right)\left( {2x - 3} \right)}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2{x^3} + {x^2} - 21x - 2{x^3} - 14{x^2} - 6x + 3{x^2} + 21x + 9}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 10{x^2} - 6x + 9}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\end{array}\)
