`x^2 + (3-2m)x - 4  = 0`

Ta có `a.c = 1.(-4) < 0`

`=>` PT có hai nghiệm trái dấu phân biệt

Vi-ét:

`{(x_1 + x_2 = 2m - 3),(x_1x_2 = -4):}`

Theo đề

`(1 - x_1^2 )(1-x_2^2) - 2x_1 - 2x_2 = -10`

`<=> 1 - x_2^2 - x_1^2 + x_1^2x_2^2 - 2(x_1 + x_2) = -10`

`<=> - x_1^2 - x_2^2 + (x_1x_2)^2 - 2(x_1 + x_2) = -11`

`<=> -(x_1^2 + x_2^2) + (x_1x_2)^2 - 2(x_1+x_2) = -11`

`<=> -[(x_1 + x_2)^2  - 2x_1x_2] + (x_1x_2)^2 - 2(x_1 + x_2) = -11`

`<=> -[(2m - 3)^2 - 2.(-4) ] + (-4)^2 - 2(2m - 3) = -11`

`<=> -(2m - 3)^2 - 8 + 16 -4m + 6 = -11`

`<=> -(2m - 3)^2 - 4m + 25 = 0`

`<=> -(4m^2 - 12m + 9) - 4m + 25 = 0`

`<=> -4m^2 + 12m - 9 - 4m + 25 = 0`

`<=> -4m^2 + 8m + 16 = 0`

`=> [(m = 1 + sqrt(5)(tm)),(m = 1 - sqrt(5)(tm)):}`
Vậy `m = 1 +- sqrt(5)`