a: \(f\left(x\right)=\left|x+2\right|-\left|x-2\right|\)

\(f\left(-x\right)=\left|-x+2\right|-\left|-x-2\right|=\left|x-2\right|-\left|x+2\right|=-f\left(x\right)\)

=>f(x) là hàm số lẻ

b: \(f\left(x\right)=\dfrac{3x^2}{2-\left|x\right|}\)

\(f\left(-x\right)=\dfrac{3\cdot\left(-x\right)^2}{2-\left|-x\right|}=\dfrac{3\cdot x^2}{2-\left|x\right|}=f\left(x\right)\)

=>f(x) là hàm số chẵn

\(\frac{x^8-1}{\left(x^4+1\right)\left(x^2-1\right)}\)

\(=\frac{\left(x^2-1\right)\left(x^4+x^2+1\right)}{\left(x^4+1\right)\left(x^2-1\right)}\)

\(=\frac{x^4+x^2+1}{x^4+1}\)

\(\frac{x^2+y^2-4+2xy}{x^2-y^2+4+4x}\)

\(=\frac{\left(x+y\right)^2-2^2}{\left(x+2\right)^2-y^2}\)

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

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

\(\frac{4x^2+12x+9}{2x^2-x-6}\)

\(=\frac{\left(2x+3\right)^2}{2x^2-4x+3x-6}\)

\(=\frac{\left(2x+3\right)^2}{2x\left(x-2\right)+3\left(x-2\right)}\)

\(=\frac{\left(2x+3\right)^2}{\left(2x+3\right)\left(x-2\right)}\)

\(=\frac{2x+3}{x-2}\)

\(\frac{25-10x+x^2}{xy-5y}\)

\(=\frac{\left(5-x\right)^2}{-y\left(5-x\right)}\)

\(=-\frac{5-x}{y}\)

\(\frac{\left|x\right|-3}{x^2-9}\)

\(=\frac{x-3}{\left(x+3\right)\left(x-3\right)}\)

\(=\frac{1}{x+3}\)

\(\frac{3\left|x-4\right|}{3x^2-3x-36}\)

\(=\frac{3\left(x-4\right)}{3\left(x^2-x-12\right)}\)

\(=\frac{x-4}{x^2-4x+3x-12}\)

\(=\frac{x-4}{x\left(x-4\right)+3\left(x-4\right)}\)

\(=\frac{x-4}{\left(x-4\right)\left(x+3\right)}\)

\(=\frac{1}{x+3}\)

Bài này dễ sáng làmcho

e: \(f\left(-x\right)=\dfrac{\left(-x\right)^4+3\cdot\left(-x\right)^2-1}{\left(-x\right)^2-4}=\dfrac{x^4+3x^2-1}{x^2-4}=f\left(x\right)\)

Vậy: f(x) là hàm số chẵn

\(c,f\left(-x\right)=\sqrt{-2x+9}=-f\left(x\right)\)

Vậy hàm số lẻ

\(d,f\left(-x\right)=\left(-x-1\right)^{2010}+\left(1-x\right)^{2010}\\ =\left[-\left(x+1\right)\right]^{2010}+\left(x-1\right)^{2010}\\ =\left(x+1\right)^{2010}+\left(x-1\right)^{2010}=f\left(x\right)\)

Vậy hàm số chẵn

\(g,f\left(-x\right)=\sqrt[3]{-5x-3}+\sqrt[3]{-5x+3}\\ =-\sqrt[3]{5x+3}-\sqrt[3]{5x-3}=-f\left(x\right)\)

Vậy hàm số lẻ

\(h,f\left(-x\right)=\sqrt{3-x}-\sqrt{3+x}=-f\left(x\right)\)

Vậy hàm số lẻ

1: \(f\left(-x\right)=\left(-x\right)^2=x^2\)

Vậy: Hàm số này chẵn

a: \(f\left(-x\right)=-2\cdot\left(-x\right)^3+3\cdot\left(-x\right)\)

\(=2x^3-3x\)

\(=-\left(-2x^3+3x\right)\)

=-f(x)

Vậy: f(x) là hàm số lẻ

c: TXĐ: D=[-2;2]

Nếu \(x\in D\Leftrightarrow-x\in D\)

\(f\left(-x\right)=\sqrt{6-3\cdot\left(-x\right)}-\sqrt{6+3\cdot\left(-x\right)}\)

\(=\sqrt{6+3x}-\sqrt{6-3x}\)

\(=-f\left(x\right)\)

Vậy: f(x) là hàm số lẻ

Câu 1: 

a) 

b) Ta có: f(x)=8

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

Vậy: Để f(x)=8 thì \(x\in\left\{2;-2\right\}\)

Ta có: \(f\left(x\right)=6-4\sqrt{2}\)

\(\Leftrightarrow2x^2=6-4\sqrt{2}\)

\(\Leftrightarrow x^2=3-2\sqrt{2}\)

\(\Leftrightarrow x=\sqrt{3-2\sqrt{2}}\)

hay \(x=\sqrt{2}-1\)

Vậy: Để \(f\left(x\right)=6-4\sqrt{2}\) thì \(x=\sqrt{2}-1\)