a) ta có : \(D=R\backslash\left\{0\right\}\) \(\Rightarrow x\in D\) thì \(-x\in D\)

ta có : \(f\left(-x\right)=\dfrac{\left(-x\right)^4+3}{\left|-x\right|+4\left(-x\right)^2}=\dfrac{x^4+3}{\left|x\right|+4x^2}=f\left(x\right)\)

\(\Rightarrow\) hàm số này là hàm chẳn.

b) ta có : \(D=R\backslash\left\{\pm1\right\}\) \(\Rightarrow x\in D\) thì \(-x\in D\)

ta có : \(f\left(-x\right)=\dfrac{3\left(-x\right)^4-\left(-x\right)^2+5}{\left|-x\right|^5-1}=\dfrac{3x^4-x^2+5}{\left|x\right|^5-1}=f\left(x\right)\)

\(\Rightarrow\) hàm số này là hàm chẳn .

c) ta có : \(D=\left(-\infty;-3\right)\cup\left(3;+\infty\right)\) \(\Rightarrow x\in D\) thì \(-x\in D\)

ta có : \(f\left(-x\right)=\dfrac{1}{\sqrt{\left(-x\right)^2-9}}=\dfrac{1}{\sqrt{x^2-9}}=f\left(x\right)\)

\(\Rightarrow\) hàm số này là hàm chẳn.

d) ta có : \(D=R\) \(\Rightarrow x\in D\) thì \(-x\in D\)

ta có : \(f\left(-x\right)=\dfrac{-x}{\left|-5x+2\right|+\left|-5x-2\right|}=\dfrac{-x}{\left|5x-2\right|+\left|5x+2\right|}=-f\left(x\right)\)

\(\Rightarrow\) hàm số này là hàm lẽ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Bài này dễ sáng làmcho

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

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

Vậy: Hàm số này chẵn

a: \(y'=\left(x^2\right)'+\left(3x\right)'-\left(6x^6\right)'+\left(\dfrac{2x-3}{x-1}\right)'\)

\(=2x+3-6\cdot6x^5+\dfrac{\left(2x-3\right)'\left(x-1\right)-\left(2x-3\right)\left(x-1\right)'}{\left(x-1\right)^2}\)

\(=-36x^5+2x+3+\dfrac{2\left(x-1\right)-2x+3}{\left(x-1\right)^2}\)

\(=-36x^5+2x+3+\dfrac{1}{\left(x-1\right)^2}\)

b: \(\left(\sqrt{2x^2-3x+1}\right)'=\dfrac{\left(2x^2-3x+1\right)'}{2\sqrt{2x^2-3x+1}}\)

\(=\dfrac{4x-3}{2\sqrt{2x^2-3x+1}}\)

\(y'=3\cdot2x-4+\dfrac{4x-3}{2\sqrt{2x^2-3x+1}}\)

\(=6x-4+\dfrac{4x-3}{2\sqrt{2x^2-3x+1}}\)

c: \(\left(\sqrt{4x^2-3x+1}\right)'=\dfrac{\left(4x^2-3x+1\right)'}{2\sqrt{4x^2-3x+1}}\)

\(=\dfrac{8x-3}{2\sqrt{4x^2-3x+1}}\)

\(y'=\left(\sqrt{4x^2-3x+1}\right)'-4'=\dfrac{8x-3}{2\sqrt{4x^2-3x+1}}\)

a.

\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)

b.

\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)

c.

\(y=\left(3x-2\right)^{\dfrac{1}{3}}\Rightarrow y'=\dfrac{1}{3}\left(3x-2\right)^{-\dfrac{2}{3}}=\dfrac{1}{3\sqrt[3]{\left(3x-2\right)^2}}\)

d.

\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)

e.

\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)

g.

\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)