Question

Tìm tập xác định:

\(y=\frac{\sqrt{2x+10-6\sqrt{2x+1}}}{\left|3x^2+5\right|x\left|-2\right|}-\frac{2017x}{\sqrt[3]{2017x-\left|2017x\right|}}\)

Xét tính chẵn lẽ của hàm số \(y=f\left(x\right)=\frac{\left|2017x-10\right|-\left|2017x+10\right|}{x^6-8x^4+16x^2}\)

Answer 1

a/ ĐKXĐ: \(\left\{{}\begin{matrix}2x+1\ge0\\3\left|x\right|^2+5\left|x\right|-2\ne0\\x-\left|x\right|\ne0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{2}\\\left|x\right|\ne\frac{1}{3}\\x< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-\frac{1}{2}\le x< 0\\x\ne-\frac{1}{3}\end{matrix}\right.\)

b/ Nếu \(x\in D\Rightarrow-x\in D\)

\(f\left(-x\right)=\frac{\left|-2017x-10\right|-\left|-2017x+10\right|}{x^6-8x^4+16x^2}\)

\(=\frac{\left|2017x+10\right|-\left|2017x-10\right|}{x^6-8x^4+16x^2}=-\frac{\left|2017x-10\right|-\left|2017x+10\right|}{x^6-8x^4+16x^2}=-f\left(x\right)\)

Hàm lẻ