`1.` Thay `m = -2` vào `(1)`. Ta có:
`x^2 - 2 ( -2 + 1 ) x + [ 2 . (-2) - 4 ] = 0`
`<=> x^2 + 2x - 8 = 0`
Ptr có: `\Delta' = 1^2 - (-8) = 9 > 0`
`=>` Ptr có `2` `n_o` pb
`x_1 = [ -b' + \sqrt{\Delta'} ] / a = 2`
`x_2 = [ -b' - \sqrt{\Delta'} ] / a = -4`
Vậy với `m = -2` thì `S = { -4 ; 2 }`
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`b)` Ptr có: `\Delta' = (m + 1)^2 - ( 2m - 4 )`
`= m^2 + 2m + 1 - 2m + 4 = m^2 + 5 > 0 AA m`
`=> \Delta' > 0 AA m`
Vậy với mọi `m` ptr luôn có `2` `n_o` pb
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`c)` Áp dụng Vi-ét. Ta có: `{(x_1 + x_2 = [-b] / a = -2m - 2),(x_1 . x_2 = c / a = 2m - 4):}`
Ta có: `A = x_1 ^2 + x_2 ^2`
`A = ( x_1 + x_2 )^2 - 2 x_1 . x_2`
`A = ( -2m - 2 )^2 - 2 ( 2m - 4 )`
`A = 4m^2 + 8m + 4 - 4m + 8`
`A = 4m^2 - 4m + 12`
1.\(m=-2\)
\(\Leftrightarrow x^2-2\left(-2+1\right)x+\left(2.-2-4\right)=0\)
\(\Leftrightarrow x^2+2x-8=0\)
\(\Delta=2^2-4.-8=4+32=36>0\)
\(\rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-2+\sqrt{36}}{2}=2\\x_2=\dfrac{-2-\sqrt{36}}{2}=-4\end{matrix}\right.\)
2.\(\Delta=\left(2\left(m+1\right)\right)^2-4\left(2m-4\right)\)
\(=4\left(m^2+2m+1\right)-4\left(2m-4\right)=4m^2+8m+4-8m+16\)
\(=4m^2+20\ge20>0\)
=> pt luôn có 2 nghiệm phân biệt
3.Theo hệ thức Vi-ét, ta có:\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1.x_2=2m-4\end{matrix}\right.\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=\left(2m+2\right)^2-2\left(2m-4\right)=4m^2+8m+4-4m+8=4m^2+4m+12\)
