Theo Vi-ét: `x_1 + x_2 = 3;x_1x_2 =1`

Ta có:

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

<=>x^2+1=3x`

`<=>x_1^2 +1=3x_1`

Ta lại có:

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

`=>x^2 = 3x-1`

`=>3x^2 =9x-3`

`=>3x_1^2 =9x_1-3`

Ta có:

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

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

`<=>x^3=3x^2-x`

`=>x_1^3 = 3x_1^2-x_1`

`=>x_1^3 = 9x_1 -3 -x_1`

`=>x_1^3= 8x_1 -3`

Ta có:

`A=|x_1^2-3x_2+1|`

`=|3x_1-3x_2|`

`=>A^2 = 9x_1^2 +9x_2^2 -18x_1x_2`

`=9(x_1+x_2)^2-36x_1x_2`

`=9.3^2-36.1`

`=45`

`=>A=sqrt45` (vi `A >= 0`)

Ta có:

`B=x_1 sqrtx_1 + sqrt(7x_2-x_1)`

`= sqrtx_1^3 + sqrt(7x_2-(3-x_2))`

`=sqrt(8x_1-3) + sqrt(8x_2-3)`

`=>B^2 = 8x_1 -3+8x_2 -3 +2sqrt((8x_1-3)(8x_2-3))`

`= 8(x_1 +x_2 )-6 +2sqrt(64x_1 x_2 -24(x_1 +x_2) +9)`

`= 8.3-6 +2sqrt(64.1 -24.3 +9)`

`= 20`

`=>B= sqrt20` (vì `B >= 0`)

`=>C= A/B = sqrt45/sqrt20=3/2`