Question

xét tính chẵn lẻ :

.y=\(\sqrt{\frac{2+x}{2-x}}\)

. y= \(x^2\left(x+\left|x\right|\right)\)

. y=\(\frac{\left|x+1\right|+\left|x-1\right|}{\sqrt{x^2+\left|x\right|}}\)

Answer 1

Giao luu:

\(y=\sqrt{\frac{2+x}{2-x}}\)

\(\left\{\begin{matrix}\left\{\begin{matrix}-2\le x< 2\\f\left(x\right)=\sqrt{\left(\frac{2+x}{2-x}\right)}\end{matrix}\right.\\\left\{\begin{matrix}-2< x\le2\\f\left(-x\right)=\sqrt{\left(\frac{2-x}{2+x}\right)}=\sqrt{\frac{1}{\left(\frac{2+x}{2-x}\right)}}=\frac{1}{\sqrt{\left(\frac{2+x}{2-x}\right)}}=\frac{1}{f\left(x\right)}\end{matrix}\right.\end{matrix}\right.\)

\(\left\{\begin{matrix}!x!< 2\\f\left(x\right).f\left(-x\right)=1\end{matrix}\right.\)

kết luân: y không chẵn không lẻ