Lời giải chi tiết:

+ TXĐ: \(D = \left[ {0;1} \right]\)

+ \(y' = \dfrac{{\left( { - 1 + 2x} \right).\left( {1 + x - {x^2}} \right) - \left( {1 - x + {x^2}} \right).\left( {1 - 2x} \right)}}{{{{\left( {1 + x - {x^2}} \right)}^2}}}\)

   \(\begin{array}{l}\,\,\,\,\,\, = \dfrac{{ - 1 - x - {x^2} + 2x + 2{x^2} - 2{x^3} - \left( {1 - 2x - x + 2{x^2} + {x^2} - 2{x^3}} \right)}}{{{{\left( {1 + x - {x^2}} \right)}^2}}}\\\,\,\,\,\,\, = \dfrac{{4x - 2}}{{{{\left( {z - x - {x^2}} \right)}^2}}}\end{array}\)

Cho \(y' = 0 \Leftrightarrow x = \dfrac{1}{2}\,\,\,\left( {tm} \right)\)

Thay \(x = 0\)vào \(f\left( x \right)\) ta có \(f\left( 0 \right) = \dfrac{1}{3}.\)

Thay \(x = \dfrac{1}{2}\)vào \(f\left( x \right)\) ta có \(f\left( {\dfrac{1}{2}} \right) = \dfrac{3}{5}.\)

Thay \(x = 1\)vào \(f\left( x \right)\) ta có \(f\left( 1 \right) = 1.\)

\( \Rightarrow \mathop {\min }\limits_{\left[ {0;1} \right]} y = \dfrac{3}{5};\mathop {\max }\limits_{\left[ {0;1} \right]} \,y = 1.\)