$$\eqalign{
& a)\,\,y = {{\sqrt {{x^2} + 1} } \over {x - 1}}\,\,D = R\backslash \left\{ 1 \right\} \cr
& \Rightarrow - 1 \in D,\,\,1 \notin D \cr
& \Rightarrow Ham\,\,so\,\,khong\,\,chan\,\,khong\,\,le. \cr
& b)\,\,y = x\left| x \right|\,\,\,D = R \cr
& y = f\left( x \right) = x\left| x \right| \cr
& f\left( { - x} \right) = - x\left| { - x} \right| = - x\left| x \right| = - f\left( x \right) \cr
& \Rightarrow Ham\,\,so\,\,la\,\,ham\,\,le \cr
& c)\,\,y = \left| {x - 1} \right|\,\,\,\,D = R \cr
& y = f\left( x \right) = \left| {x - 1} \right| \cr
& f\left( { - x} \right) = \left| { - x - 1} \right| = \left| {x + 1} \right| \cr
& \Rightarrow Ham\,\,so\,\,khong\,\,chan\,\,khong\,\,le. \cr
& d)\,\,y = {x^2} + 2\left| x \right| + 2 \cr
& D = R \cr
& y = f\left( x \right) = {x^2} + 2\left| x \right| + 2 \cr
& f\left( { - x} \right) = {\left( { - x} \right)^2} + 2\left| { - x} \right| + 2 \cr
& f\left( { - x} \right) = {x^2} + 2\left| x \right| + 2 = f\left( x \right) \cr
& \Rightarrow Ham\,\,so\,\,la\,\,ham\,\,chan \cr} $$
