Giải thích các bước giải:
`1,`
`x^3-3x^2+4=0`
`<=> x^3+x^2-4x^2+4=0`
`<=> x^2(x+1)-4(x^2-1)=0`
`<=> x^2(x+1)-4(x-1)(x+1)=0`
`<=> (x+1)(x^2-4x+4)=0`
`<=> (x+1)(x-2)^2=0`
`<=> [(x+1=0),((x-2)^2=0):} <=> [(x=-1),(x=2):}`
Vậy `x in {-1;2}`
`2,`
`x^4+x^3-4x^2+5x-3=0`
`<=> x^4-x^3+2x^3-2x^2-2x^2+2x+3x-3=0`
`<=> x^3(x-1)+2x^2(x-1)-2x(x-1)+3(x-1)=0`
`<=> (x-1)(x^3+2x^2-2x+3)=0`
`<=> (x-1)(x^3+3x^2-x^2-3x+x+3)=0`
`<=> (x-1)[x^2(x+3)-x(x+3)+(x+3)]=0`
`<=> (x-1)(x^2-x+1)(x+3)=0`
Mà : `x^2-x+1=(x-1/2)^2+3/4>=3/4>0`
`to [(x-1=0),(x+3=0):} <=> [(x=1),(x=-3):}`
Vậy `x in {-3;1}`
