\(\text{Δ}=\left(-2m\right)^2-4\left(m-1\right)\)
\(=4m^2-4m+4\)
\(=4m^2-4m+1+3=\left(2m-1\right)^2+3>=3>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)
\(x_1^2\cdot x_2+m\cdot x_2-x_2=4\)
=>\(x_1^2\cdot x_2+x_2\left(m-1\right)=4\)
=>\(x_1^2\cdot x_2+x_2\cdot x_1x_2=4\)
=>\(x_1x_2\left(x_1+x_2\right)=4\)
=>\(2m\left(m-1\right)=4\)
=>m(m-1)=2
=>\(m^2-m-2=0\)
=>(m-2)(m+1)=0
=>\(\left[{}\begin{matrix}m=2\\m=-1\end{matrix}\right.\)
Δ = ( − 2 m ) 2 − 4 ( m − 1 ) = 4 m 2 − 4 m + 4 = 4 m 2 − 4 m + 1 + 3 = ( 2 m − 1 ) 2 + 3 >= 3 > 0 ∀ m =>Phương trình luôn có hai nghiệm phân biệt Theo Vi-et, ta có: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ x 1 + x 2 = − b a = 2 m x 1 x 2 = c a = m − 1 x 2 1 ⋅ x 2 + m ⋅ x 2 − x 2 = 4 => x 2 1 ⋅ x 2 + x 2 ( m − 1 ) = 4 => x 2 1 ⋅ x 2 + x 2 ⋅ x 1 x 2 = 4 => x 1 x 2 ( x 1 + x 2 ) = 4 => 2 m ( m − 1 ) = 4 =>m(m-1)=2 => m 2 − m − 2 = 0 =>(m-2)(m+1)=0 => [ m = 2 m = − 1
