Câu 3:x^4-4x^3+6x^2-4x+căn bậc hai của x^2-2x+17=3

Câu 3:

`x^4-4x^3+6x^2-4x+sqrt(x^2-2x+17)=3`

`<=>x^4-4x^3+6x^2-4x-3+sqrt(x^2-2x+17)=0`

`<=>x^4-4x^3+6x^2-4x+1+(sqrt(x^2-2x+17)-4)=0`

`<=>(x^4-2x^3+x^2)+(-2x^3+4x^2-2x)+(x^2-2x+1)`

`+((sqrt(x^2-2x+17)-4)(sqrt(x^2-2x+17)+4))/(sqrt(x^2-2x+17)+4)=0`

`<=>x^2(x^2-2x+1)-2x(x^2-2x+1)+(x^2-2x+1)`

`+((sqrt(x^2-2x+17))^2-4^2)/(sqrt(x^2-2x+17)+4)=0`

`<=>(x^2-2x+1)(x^2-2x+1)+(x^2-2x+17-16)/(sqrt(x^2-2x+17)+4)=0`

`<=>(x^2-2x+1)^2+(x^2-2x+1)/(sqrt(x^2-2x+17)+4)=0`

`<=>[(x-1)^2]^2+((x-1)^2)/(sqrt(x^2-2x+17)+4)=0`

`<=>(x-1)^2 [(x-1)^2+1/(sqrt(x^2-2x+17)+4)]=0`

Vì `1/(sqrt(x^2-2x+17)+4)>0` `AA x`

`(x-1)^2>=0` `AA x`

`=>(x-1)^2+1/(sqrt(x^2-2x+17)+4)>0` `AA x`

`=>(x-1)^2=0`

`<=>x=1`

Vậy `S={1}`

`- \text{Lamtoanbangcatinhmang} -`