Question

xét tính chẵn lẻ của hàm số y = f(x) =\(\dfrac{x^2+1}{|2x + 1|+|2x - 1|}\)

Answer 1

\(TXD\) \(D=R\backslash\left\{0\right\}\) là tập đối xứng.

\(\forall x\in D\Rightarrow-x\in D\)

Có \(f\left(-x\right)=\dfrac{\left(-x\right)^2+1}{\left|2\left(-x\right)+1\right|+\left|2\left(-x\right)-1\right|}\)

\(=\dfrac{x^2+1}{\left|1-2x\right|+\left|-2x-1\right|}\)

\(=\dfrac{x^2+1}{\left|-\left(2x-1\right)\right|+\left|-\left(2x+1\right)\right|}\)

\(=\dfrac{x^2+1}{\left|2x-1\right|+\left|2x+1\right|}\) \(=f\left(x\right)\)

Vậy hàm số \(y=f\left(x\right)=\dfrac{x^2+1}{\left|2x+1\right|+\left|2x-1\right|}\) là hàm số chẵn.

Answer 2

TXĐ: D=R

Khi \(x\in D\) thì \(-x\in D\)

\(f\left(-x\right)=\dfrac{\left(-x\right)^2+1}{\left|-2x+1\right|+\left|-2x-1\right|}\)

\(=\dfrac{x^2+1}{\left|2x+1\right|+\left|2x-1\right|}=f\left(x\right)\)

=>f(x) chẵn