Đáp án:
a) Chẵn
b) Chẵn
c) Không chẵn không lẻ.
Giải thích các bước giải:
\(\eqalign{
& a)\,\,A = {x^4} - 2{x^2} + 2 = f\left( x \right) \cr
& TXD:\,\,D = R \Rightarrow \forall x \in D \Rightarrow - x \in D \cr
& f\left( { - x} \right) = {\left( { - x} \right)^4} - 2{\left( { - x} \right)^2} + 2 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^4} - 2{x^2} + 2 = f\left( x \right) \cr
& \Rightarrow Ham\,\,so\,\,la\,\,ham\,\,chan. \cr
& b)\,\,B = \left| x \right| + 1 = f\left( x \right) \cr
& TXD:\,\,D = R \Rightarrow \forall x \in D \Rightarrow - x \in D \cr
& f\left( { - x} \right) = \left| { - x} \right| + 1 = \left| x \right| + 1 = f\left( x \right) \cr
& \Rightarrow Ham\,\,so\,\,la\,\,ham\,\,chan. \cr
& c)\,\,y = {\left( {x - 2} \right)^2} \cr
& TXD:\,\,D = R \Rightarrow \forall x \in D \Rightarrow - x \in D \cr
& f\left( { - x} \right) = {\left( { - x - 2} \right)^2} = {\left( {x + 2} \right)^2} \ne \pm f\left( x \right) \cr
& \Rightarrow Ham\,\,so\,\,khong\,\,chan\,\,khong\,\,le. \cr} \)
