a, \(f\left(x\right)=2x^4-x^3+4x^2-x\)

\(\Rightarrow f'\left(x\right)=\left(2x^4-x^3+4x^2-x\right)'\)

\(=\left(2x^4\right)'-\left(x^3\right)'+\left(4x^2\right)'-\left(x\right)'\)

\(=2.4x^3-3x^2+4.2x-1\)

\(=8x^3-3x^2+8x-1\)

b, \(f\left(x\right)=2sinx\)

\(\Rightarrow f'\left(x\right)=\left(2sinx\right)'=2cosx\)

c, \(f\left(x\right)=\dfrac{3x^2+2x-5}{x}\)

\(\Rightarrow f'\left(x\right)=\left(\dfrac{3x^2+2x-5}{x}\right)'\)

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

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

\(=3+0+\dfrac{5}{x^2}=\dfrac{5}{x^2}+3\)

b a d a c a d b c d b d d b a c b a a a

Bài 3:

a:Ta có: \(\sqrt{4x^2-4x+1}=3\)

\(\Leftrightarrow\left|2x-1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b: Ta có: \(2\sqrt{x-1}-4=0\)

\(\Leftrightarrow x-1=4\)

hay x=5

h: \(\dfrac{x^3+8}{x+2}=x^2-2x+4\)

i: \(\dfrac{27x^3-1}{9x^2+3x+1}=3x-1\)

dài quá bạn ơi

pls :((

Bài 3: 

a: 11/120=11/120

7/40=21/120

c: 24/146=312/1898

6/13=876/1898

c: 7/30=28/120

-9/40=-27/120

13/60=26/120

d: 17/60=51/180

-5/18=-50/180

-64/90=-128/180