Δ=[−2(m+1)]2−4⋅1(m2+2m)

\(= 4 \left(\right. m^{2} + 2 m + 1 \left.\right) - 4 \left(\right. m^{2} + 2 m \left.\right) = 4 > 0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt là:

\(\left[\right. x = \frac{\left(\right. 2 m + 2 \left.\right) - \sqrt{4}}{2} = \frac{2 m + 2 - 2}{2} = m \\ x = \frac{2 m + 2 + 2}{2} = \frac{2 m + 4}{2} = m + 2\)

Vì m<m+2 nên \(x_{1} = m ; x_{2} = m + 2\)

\(\mid x_{1} \mid = 3 \mid x_{2} \mid\)

=>\(\mid m \mid = 3 \mid m + 2 \mid = \mid 3 m + 6 \mid\)

=>\(\left[\right. 3 m + 6 = m \\ 3 m + 6 = - m \Leftrightarrow \left[\right. 2 m = - 6 \\ 4 m = - 6 \Leftrightarrow \left[\right. m = - 3 \\ m = - \frac{3}{2}\)

Δ=[−2(m+1)]2−4⋅1(m2+2m)

\(= 4 \left(\right. m^{2} + 2 m + 1 \left.\right) - 4 \left(\right. m^{2} + 2 m \left.\right) = 4 > 0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt là:

\(\left[\right. x = \frac{\left(\right. 2 m + 2 \left.\right) - \sqrt{4}}{2} = \frac{2 m + 2 - 2}{2} = m \\ x = \frac{2 m + 2 + 2}{2} = \frac{2 m + 4}{2} = m + 2\)

Vì m<m+2 nên \(x_{1} = m ; x_{2} = m + 2\)

\(\mid x_{1} \mid = 3 \mid x_{2} \mid\)

=>\(\mid m \mid = 3 \mid m + 2 \mid = \mid 3 m + 6 \mid\)

=>\(\left[\right. 3 m + 6 = m \\ 3 m + 6 = - m \Leftrightarrow \left[\right. 2 m = - 6 \\ 4 m = - 6 \Leftrightarrow \left[\right. m = - 3 \\ m = - \frac{3}{2}\)

Δ=[−2(m+1)]2−4⋅1(m2+2m)

\(= 4 \left(\right. m^{2} + 2 m + 1 \left.\right) - 4 \left(\right. m^{2} + 2 m \left.\right) = 4 > 0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt là:

\(\left[\right. x = \frac{\left(\right. 2 m + 2 \left.\right) - \sqrt{4}}{2} = \frac{2 m + 2 - 2}{2} = m \\ x = \frac{2 m + 2 + 2}{2} = \frac{2 m + 4}{2} = m + 2\)

Vì m<m+2 nên \(x_{1} = m ; x_{2} = m + 2\)

\(\mid x_{1} \mid = 3 \mid x_{2} \mid\)

=>\(\mid m \mid = 3 \mid m + 2 \mid = \mid 3 m + 6 \mid\)

=>\(\left[\right. 3 m + 6 = m \\ 3 m + 6 = - m \Leftrightarrow \left[\right. 2 m = - 6 \\ 4 m = - 6 \Leftrightarrow \left[\right. m = - 3 \\ m = - \frac{3}{2}\)

Δ=[−2(m+1)]2−4⋅1(m2+2m)

\(= 4 \left(\right. m^{2} + 2 m + 1 \left.\right) - 4 \left(\right. m^{2} + 2 m \left.\right) = 4 > 0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt là:

\(\left[\right. x = \frac{\left(\right. 2 m + 2 \left.\right) - \sqrt{4}}{2} = \frac{2 m + 2 - 2}{2} = m \\ x = \frac{2 m + 2 + 2}{2} = \frac{2 m + 4}{2} = m + 2\)

Vì m<m+2 nên \(x_{1} = m ; x_{2} = m + 2\)

\(\mid x_{1} \mid = 3 \mid x_{2} \mid\)

=>\(\mid m \mid = 3 \mid m + 2 \mid = \mid 3 m + 6 \mid\)

=>\(\left[\right. 3 m + 6 = m \\ 3 m + 6 = - m \Leftrightarrow \left[\right. 2 m = - 6 \\ 4 m = - 6 \Leftrightarrow \left[\right. m = - 3 \\ m = - \frac{3}{2}\)

Δ=[−2(m+1)]2−4⋅1(m2+2m)

\(= 4 \left(\right. m^{2} + 2 m + 1 \left.\right) - 4 \left(\right. m^{2} + 2 m \left.\right) = 4 > 0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt là:

\(\left[\right. x = \frac{\left(\right. 2 m + 2 \left.\right) - \sqrt{4}}{2} = \frac{2 m + 2 - 2}{2} = m \\ x = \frac{2 m + 2 + 2}{2} = \frac{2 m + 4}{2} = m + 2\)

Vì m<m+2 nên \(x_{1} = m ; x_{2} = m + 2\)

\(\mid x_{1} \mid = 3 \mid x_{2} \mid\)

=>\(\mid m \mid = 3 \mid m + 2 \mid = \mid 3 m + 6 \mid\)

=>\(\left[\right. 3 m + 6 = m \\ 3 m + 6 = - m \Leftrightarrow \left[\right. 2 m = - 6 \\ 4 m = - 6 \Leftrightarrow \left[\right. m = - 3 \\ m = - \frac{3}{2}\)

Δ=[−2(m+1)]2−4⋅1(m2+2m)

\(= 4 \left(\right. m^{2} + 2 m + 1 \left.\right) - 4 \left(\right. m^{2} + 2 m \left.\right) = 4 > 0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt là:

\(\left[\right. x = \frac{\left(\right. 2 m + 2 \left.\right) - \sqrt{4}}{2} = \frac{2 m + 2 - 2}{2} = m \\ x = \frac{2 m + 2 + 2}{2} = \frac{2 m + 4}{2} = m + 2\)

Vì m<m+2 nên \(x_{1} = m ; x_{2} = m + 2\)

\(\mid x_{1} \mid = 3 \mid x_{2} \mid\)

=>\(\mid m \mid = 3 \mid m + 2 \mid = \mid 3 m + 6 \mid\)

=>\(\left[\right. 3 m + 6 = m \\ 3 m + 6 = - m \Leftrightarrow \left[\right. 2 m = - 6 \\ 4 m = - 6 \Leftrightarrow \left[\right. m = - 3 \\ m = - \frac{3}{2}\)

Δ=(−m)2−4(m−2)

\(= m^{2} - 4 m + 8 = \left(\left(\right. m - 2 \left.\right)\right)^{2} + 4 > = 4 > 0 \forall m\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\({x_1+x_2=-\frac{b}{a}=m,x_1x_2=\frac{c}{a}=m-2}\)

\(x_{1} - x_{2} = 2 \sqrt{5}\)

=>\(\left(\left(\right. x_{1} - x_{2} \left.\right)\right)^{2} = \left(\left(\right. 2 \sqrt{5} \left.\right)\right)^{2} = 20\)

=>\(\left(\left(\right. x_{1} + x_{2} \left.\right)\right)^{2} - 4 x_{1} x_{2} = 20\)

=>\(m^{2} - 4 \left(\right. m - 2 \left.\right) = 20\)

=>\(m^{2} - 4 m - 12 = 0\)

=>(m-6)(m+2)=0

=>\(\left[\right. m - 6 = 0 \\ m + 2 = 0 \Leftrightarrow \left[\right. m = 6 \\ m = - 2\)

Δ=(−m)2−4(m−2)

\(= m^{2} - 4 m + 8 = \left(\left(\right. m - 2 \left.\right)\right)^{2} + 4 > = 4 > 0 \forall m\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\({x_1+x_2=-\frac{b}{a}=m,x_1x_2=\frac{c}{a}=m-2}\)

\(x_{1} - x_{2} = 2 \sqrt{5}\)

=>\(\left(\left(\right. x_{1} - x_{2} \left.\right)\right)^{2} = \left(\left(\right. 2 \sqrt{5} \left.\right)\right)^{2} = 20\)

=>\(\left(\left(\right. x_{1} + x_{2} \left.\right)\right)^{2} - 4 x_{1} x_{2} = 20\)

=>\(m^{2} - 4 \left(\right. m - 2 \left.\right) = 20\)

=>\(m^{2} - 4 m - 12 = 0\)

=>(m-6)(m+2)=0

=>\(\left[\right. m - 6 = 0 \\ m + 2 = 0 \Leftrightarrow \left[\right. m = 6 \\ m = - 2\)

Δ=(−2)2−4(m−1)=4−4m+4=−4m+8

Để phương trình có hai nghiệm phân biệt thì Δ>=0

=>-4m+8>=0

=>-4m>=-8

=>m<=2

Theo Vi-et, ta có:

\({x_1+x_2=-\frac{b}{a}=2,x_1x_2=\frac{c}{a}=m-1}\)

\(x_{1}^{2} + x_{2}^{2} - x_{1} x_{2} + x_{1}^{2} \cdot x_{2}^{2} - 14 = 0\)

=>\(\left(\left(\right. x_{1} + x_{2} \left.\right)\right)^{2} - 3 x_{1} x_{2} + \left(\left(\right. x_{1} x_{2} \left.\right)\right)^{2} - 14 = 0\)

=>\(2^{2} - 3 \left(\right. m - 1 \left.\right) + \left(\left(\right. m - 1 \left.\right)\right)^{2} - 14 = 0\)

=>\(4 - 3 m + 3 + m^{2} - 2 m + 1 - 14 = 0\)

=>\(m^{2} - 5 m - 6 = 0\)

=>(m-6)(m+1)=0

=>\(\left[\right.m=6\left(\right.loại\left.\right)\\m=-1\left(\right.nhận\left.\right)\)

Δ=(−2)2−4(m−1)=4−4m+4=−4m+8

Để phương trình có hai nghiệm phân biệt thì Δ>=0

=>-4m+8>=0

=>-4m>=-8

=>m<=2

Theo Vi-et, ta có:

\({x_1+x_2=-\frac{b}{a}=2,x_1x_2=\frac{c}{a}=m-1}\)

\(x_{1}^{2} + x_{2}^{2} - x_{1} x_{2} + x_{1}^{2} \cdot x_{2}^{2} - 14 = 0\)

=>\(\left(\left(\right. x_{1} + x_{2} \left.\right)\right)^{2} - 3 x_{1} x_{2} + \left(\left(\right. x_{1} x_{2} \left.\right)\right)^{2} - 14 = 0\)

=>\(2^{2} - 3 \left(\right. m - 1 \left.\right) + \left(\left(\right. m - 1 \left.\right)\right)^{2} - 14 = 0\)

=>\(4 - 3 m + 3 + m^{2} - 2 m + 1 - 14 = 0\)

=>\(m^{2} - 5 m - 6 = 0\)

=>(m-6)(m+1)=0

=>\(\left[\right.m=6\left(\right.loại\left.\right)\\m=-1\left(\right.nhận\left.\right)\)