Question

Answer 1

`a)` Thay `m=2` vào `(1)` có:

   `x^2-(2.2+1)x+2^2+2-2=0`

`<=>x^2-5x+4=0`

`<=>(x-4)(x-1)=0`

`<=>[(x=4),(x=1):}`

`b)` Ptr có `2` nghiệm phân biệt

   `<=>\Delta > 0`

   `<=>(2m+1)^2-4(m^2+m-2) > 0`

   `<=>4m^2+4m+1-4m^2-4m+8 > 0`

    `<=>9 > 0` (LĐ `AA m`)

`=>` Áp dụng Viét có: `{(x_1+x_2=-b/a=2m+1),(x_1 .x_2=c/a=m^2+m-2):}`

Ta có: `x_1(x_1-2x_2)+x_2(x_2-3x_1)=9`

`<=>x_1 ^2-2x_1 .x_2+x_2 ^2-3x_1 .x_2=9`

`<=>(x_1 +x_2)^2-7x_1 .x_2=9`

`<=>(2m+1)^2-7(m^2+m-2)=9`

`<=>4m^2+4m+1-7m^2-7m+14-9=0`

`<=>-3m^2-3m+6=0`

`<=>m^2+m-2=0`

Ptr có: `a+b+c=1+1-2=0` 

  `=>m_1=1;m_2=c/a=-2`