Question

5. \(x_1+x_2=2m+2;x_1x_2=m^2+4\)

tìm m để \(x_1^2+2\left(m+1\right)x_2< =3m^2+16\)

 

Answer 1

Ta có: `x_1 ^2+2(m+1)x_2 <= 3m^2+16`

 `<=>x_1^2+(x_1+x_2)x_2 <= 3m^2+16`

`<=>(x_1+x_2)^2-x_1 .x_2 <= 3m^2+16`

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

`<=>8m <= 16`

`<=>m <= 2`.