Đáp án:

`A = 22`
`B = 727`
`C = −140`
`D =` $\sqrt{29}$

Giải thích:

PT: `x^2 − 5x − 1 = 0`  `(1)`

`Δ = b^2 − 4ac = (−5)^2 − 4*1*(−1) = 29 > 0`

`⇒` Phương trình `(1)` có `2` nghiệm phân biệt `x_1, x_2`

Theo Vi-ét: `x_1 + x_2 = 5; x_1x_2 = −1`

`A = x_1^2 + x_2^2 − x_1 − x_2`

`= (x_1 + x_2)^2 − 2x_1x_2 − (x_1+ x_2)`

`= 25 − 2(−1) − 5`

`= 22`

Vậy `A = 22`

`B = x_1^4 + x_2^4` 

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

Mà `x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = 5^2 - 2(-1) = 27`

`B = 27^2 - 2(-1)^2`

`= 729 - 2`

`= 727`
Vậy `B = 727`

`C =1/(x_1^3)+1/(x_2^3)`

`= (x_1^3 + x_2^3) / (x_1x_2)^3`

`= [(x_1 + x_2)^3 - 3x_1x_2(x_1 + x_2)] / (x_1x_2)^3`

`= [5^3 - 3(-1)(5)] / (-1)^3`

`= (125 + 15) / -1`

`= -140`

Vậy `C = -140`

`D = |x_1 − x_2|`

$=\sqrt{(x_1 + x_2)^2 − 4x_1x_2}$

$=\sqrt{25 − 4(−1)}$

$=\sqrt{29}$

Vậy $D=\sqrt{29}$