Đáp án:
a) lẻ
b) chẵn
c) lẻ
d) chẵn
e) chẵn
f) chẵn
Giải thích các bước giải:
\(\begin{array}{l}
a)y\left( { - x} \right) = \dfrac{{\cos \left( { - 2x} \right)}}{{ - x}} = - \dfrac{{\cos 2x}}{x} = - y\left( x \right)\\
\to lẻ\\
d)y\left( { - x} \right) = {\tan ^7}\left( { - 2x} \right).\sin \left( { - 5x} \right)\\
= {\tan ^7}2x.\sin 5x = y\left( x \right)\\
\to chẵn\\
b)y\left( { - x} \right) = \sqrt {1 - \cos \left( { - x} \right)} = \sqrt {1 - \cos x} = y\left( x \right)\\
\to chẵn\\
e)y\left( { - x} \right) = {\left( { - x} \right)^2} + {\tan ^2}\left| { - x} \right|\\
= {x^2} + {\tan ^2}\left| x \right| = y\left( x \right)\\
\to chẵn\\
c)y\left( { - x} \right) = \sin \left( { - x} \right).\cos \left( { - x} \right) + \tan \left( { - x} \right)\\
= - \sin x.\cos x - \tan x = - \left( {\sin x.\cos x + \tan x} \right)\\
= - y\left( x \right)\\
\to lẻ\\
f)y\left( { - x} \right) = \dfrac{{{{\sin }^4}\left( { - x} \right) + 1}}{{2 + {{\cos }^6}\left( { - x} \right)}} = \dfrac{{{{\sin }^4}x + 1}}{{2 + {{\cos }^6}x}} = y\left( x \right)\\
\to chẵn
\end{array}\)
