a: TXĐ: D=R\{0}Khi \(x\in D\) thì \(-x\in D\)\(f\left(-x\right)=\dfrac{sin\left(-3x\right)}{-x}=\dfrac{-sin3x}{-x}=\dfrac{sin3x}{x}=f\left(x\right)\)=>f(x) chẵnb: ĐKXĐ: x<>kpi và sin x<>0=>x<>kpi=>D=R\{kpi}\(x\in D\Leftrightarrow-x\in D\)\(f\left(-x\right)=cot\left(-x\right)-\dfrac{2}{sin\left(-x\right)}\)\(=-cotx+\dfrac{2}{sinx}=-f\left(x\right)\)=>f(x) lẻCâu trả lời: a: TXĐ: D=R\{0}Khi \(x\in D\) thì \(-x\in D\)\(f\left(-x\right)=\dfrac{sin\left(-3x\right)}{-x}=\dfrac{-sin3x}{-x}=\dfrac{sin3x}{x}=f\left(x\right)\)=>f(x) chẵnb: ĐKXĐ: x<>kpi và sin x<>0=>x<>kpi=>D=R\{kpi}\(x\in D\Leftrightarrow-x\in D\)\(f\left(-x\right)=cot\left(-x\right)-\dfrac{2}{sin\left(-x\right)}\)\(=-cotx+\dfrac{2}{sinx}=-f\left(x\right)\)=>f(x) lẻ