Khảo sát tính chẵn, lẻ của hàm số:
1) \(y=3x^{2}-5\)                                       2) \(y=\sqrt{1+x}\)
1)  \(y=f(x)=3x^{2}-5\) có miền xác định \(D=R\) là tập đối xứng
và \(f(-x)=3(-x)^{2}-5=3x^{2}-5=f(x)\)
Do đó $f(x)=f(-x)  \forall x\in D.$
Vậy \(y=f(x)=3x^{2}-5\) là hàm số chẵn.

2)  \(y=f(x)=\sqrt{1+x}\) có miền xác định \(D=[-1,+\infty )\)
Do D không phải là 1 tập đối xứng, nên ta không xét tính chẵn lẻ của hàm số.
Hàm số này không chẵn, không lẻ

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