`1)x^4+x^3+4x^2+5x+25=0`

`(x^4-2x^3+5x^2)+(3x^3-6x^2+15x)+(5x^2-10x+25)=0`

`x^2(x^2-2x+5)+3x(x^2-2x+5)+5(x^2-2x+5)=0`

`(x^2+3x+5)(x^2-2x+5)=0`

`TH1:x^2+3x+5=0`

`(x^2+3x+9/4)+11/4=0`

`[x^2+2*x*3/2+(3/2)^2]+11/4=0`

`(x+3/2)^2=-11/4` (vô lý)

`TH2:x^2-2x+5=0`

`(x^2-2x+1)+4=0`

`(x-1)^2=-4` (vô lý)

Vậy phương trình vô nghiệm

`2)x^4+3x^3-14x^2-6x+4=0`

`(x^4-2x^3-2x^2)+(5x^3-10x^2-10x)+(-2x^2+4x+4)=0`

`x^2(x^2-2x-2)+5x(x^2-2x-2)-2(x^2-2x-2)=0`

`(x^2-2x-2)(x^2+5x-2)=0`

`TH1:x^2-2x-2=0`

`(x^2-2x+1)-3=0`

`(x-1)^2=3`

`x-1=+-\sqrt{3}`

`x=1+-\sqrt{3}`

`TH2:x^2+5x-2=0`

`(x^2+5x+25/4)-33/4=0`

`(x+5/2)^2=33/4`

`x+5/2=+-\sqrt{33}/2`

`x=+-\sqrt{33}/2-5/2`

Vậy: `S={1-\sqrt{3};1+\sqrt{3};\sqrt{33}/2-5/2;-\sqrt{33}/2-5/2}`