`41.C `

`pt:x^2-2x+3-m=0 `

`Δ=(-2)^2-4.1.(3-m) `

`=4-12+4m `

`=4m-8 `

Để pt có `2 ` nghiệm thì `Δ=4m-8>=0 `

`4m>=8 `

`m>=2 `

Áp dụng định lí Viète, có`: `

`{(x_{1}+x_{2}=2(1)),(x_{1}x_{2}=3-m(2)):}`

`S=(x_{1}^2-2_{1})(x_{2}^2-2x_{2})+4m-8 `

`=x_{1}^2x_{2}^2-2x_{1}^2x_{2}-2x_{1}x_{2}^2+4x_{1}x_{2}+4m-8 `

`=x_{1}x_{2}(x_{1}x_{2})-2x_{1}x_{2}(x_{1}+x_{2})+4x_{1}x_{2}+4m-8(3) `

Thay `(1),(2) ` vào `(3), ` có`: `

`S=(3-m)(3-m)-2(3-m)2+4(3-m)+4m-8 `

`=(3-m)^2-4(3-m)+12-4m+4m-8 `

`=(3-m)^2-12+4m+12-4m+4m-8 `

`=9-6m+m^2+4m-8 `

`=m^2-2m+1 `

`=(m-1)^2>=0(L Đ) `

mà `m-1=0 `

`m=1(KTM) ` 

`Đk:m>=2->m=2 `

Vậy `M i n _{S}=(2-1)^2=1^2=1 ` khi `m=2 `