Em xem lại đề câu c nhé \(\begin{array}{l} a)\dfrac{{ - 4{x^2} + 4x}}{{15}} = \dfrac{{1 - \left( {4{x^2} - 4x + 1} \right)}}{{15}} = \dfrac{{1 - {{\left( {2x - 1} \right)}^2}}}{{15}} \le \dfrac{1}{{15}}\\ GTLN\,la\,\dfrac{1}{{15}} \Leftrightarrow 2x - 1 = 0 \Leftrightarrow x = \dfrac{1}{2}\\ b)\dfrac{{5 - \left| {1 - 4x} \right|}}{{12}} \le \dfrac{5}{{12}}\,\,\left( {do\,\,\left| {1 - 4x} \right| \ge 0} \right)\\ GTLN\,\,la\,\,\dfrac{5}{{12}} \Leftrightarrow 1 - 4x = 0 \Leftrightarrow x = \dfrac{1}{4}\\ c)\,\dfrac{{9 - {{\left( {x + 5} \right)}^2}}}{3} \le \dfrac{9}{3} \Leftrightarrow \dfrac{{9 - {{\left( {x + 5} \right)}^2}}}{3} \le 3\,\left( {do\,\,{{\left( {x + 5} \right)}^2} \ge 0} \right)\\ GTLN\,la\,\,3 \Leftrightarrow x + 5 = 0 \Leftrightarrow x = - 5 \end{array}\)
