`a)` Đúng

`f(x)=2x+1`

`-> f'(x)=(2x)'+1'=2`

`g(x)=x^2-x+2`

`-> g'(x)=(x^2)'-x'+2'=2x-1`

`b)` Sai

`k(x)=[g(x)]/[f(x)]=[x^2-x+2]/[2x+1]`

`-> k'(x)=[(x^2-x+2)'(2x+1)-(2x+1)'(x^2-x+2)]/[(2x+1)^2]`

`->k'(x)=[(2x-1)(2x+1)-2(x^2-x+2)]/[(2x+1)^2]`

`-> k'(x)=[4x^2-1-2x^2+2x-4]/[(2x+1)^2]`

`-> k'(x)=[2x^2+2x-5]/[(2x+1)^2]`

 `c)` Sai

`h(x)=f(x)g(x)=(2x+1)(x^2-x+2)`

`-> h'(x)=(2x+1)'(x^2-x+2)+(2x+1)(x^2-x+2)'`

`-> h'(x)=2(x^2-x+2)+(2x+1)(2x-1)`

`-> h'(x)=2x^2-2x+4+4x^2-1`

`-> h'(x)=6x^2-2x+3`

`d)` Sai

`h'(x)=0`

`-> 6x^2-2x+3=0`

`-> x^2-1/3x+1/2=0`

`-> x^2-1/3x+1/36+7/18=0`

`-> (x-1/6)^2+7/18=0` (vô lí)

`h'(x)=0` vô nghiệm