TXĐ: \(D=\mathbb R\)\(\eqalign{
& f\left( x \right) = {\left( {{{\sin }^2}x} \right)^2} + {\left( {{{\cos }^2}x} \right)^2} + 2{\sin ^2}x{\cos ^2}x - 2{\sin ^2}x{\cos ^2}x \cr
& \,\,\,\,\,\,\,\,\,\,\, = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x = 1 - {1 \over 2}{\sin ^2}2x \cr} \)Vì \(0 \le {\sin ^2}2x \le 1\) nên: \(\,\,f\left( x \right) \le 1\) với mọi \(x \in {\mathbb{R}},f\left( 0 \right) = 1\). Vậy \(\mathop {\max f\left( x \right)}\limits_{x \in {\mathbb {R}}} = 1\)\(*\,\,\,f\left( x \right) \ge {1 \over 2}\) với mọi \(x \in {\mathbb{R}},f\left( {{\pi \over 4}} \right) = 1 - {1 \over 2} = {1 \over 2}\)Vậy \(\mathop {\min f\left( x \right)}\limits_{x \in {\mathbb {R}}} = {1 \over 2}\).
