Δ=(−m)2−4(m2−m−3)

\(= m^{2} - 4 m^{2} + 4 m + 12 = - 3 m^{2} + 4 m + 12\)

Để phương trình có hai nghiệm thì Δ>=0

=>\(- 3 m^{2} + 4 m + 12 > = 0\)

=>\(3 m^{2} - 4 m - 12 < = 0\)

=>\(\frac{2 - 2 \sqrt{10}}{3} < = m < = \frac{2 + 2 \sqrt{10}}{3}\)

Theo Vi-et, ta có:

\(\left[\begin{array}{l}x_1+x_2=-\frac{b}{a}=m\\ x_1x_2=\frac{c}{a}=m^2-m-3\end{array}\right.\)

\(x_{1} ; x_{2}\) là độ dài các cạnh góc vuông của một tam giác vuông ABC có độ dài cạnh huyền là 2 nên ta có:

\(x_{1}^{2} + x_{2}^{2} = 2^{2}\)

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

=>\(m^{2} - 2 \left(\right. m^{2} - m - 3 \left.\right) = 4\)

=>\(m^{2} - 2 m^{2} + 2 m + 6 = 4\)

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

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

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

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

=>\(\left[\right. m - 1 = \sqrt{3} \\ m - 1 = - \sqrt{3} \Leftrightarrow \left[\right. m = \sqrt{3} + 1 \left(\right. n h ậ n \left.\right) \\ m = - \sqrt{3} + 1 \left(\right. n h ậ n \left.\right)\)

Δ=(−2)2−4⋅1⋅(m−1)=4−4m+4=−4m+8

Để phương trình có hai nghiệm thì Δ>=0

=>\(- 4 m + 8 > = 0\)

=>-4m>=-8

=>m<=2

Theo vi-et, ta có:

\(\left{\right. x_{1} + x_{2} = - \frac{b}{a} = 2 \\ x_{1} x_{2} = \frac{c}{a} = m - 1\)

\(x_{1}^{4} - x_{1}^{3} = x_{2}^{4} - x_{2}^{3}\)

=>\(x_{1}^{4} - x_{2}^{4} = x_{1}^{3} - x_{2}^{3}\)

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

=>\(\left(\right. x_{1} - x_{2} \left.\right) \cdot 2 \cdot \left[\right. \left(\left(\right. x_{1} + x_{2} \left.\right)\right)^{2} - 2 x_{1} x_{2} \left]\right. = \left(\right. x_{1} - x_{2} \left.\right) \cdot \left[\right. \left(\left(\right. x_{1} + x_{2} \left.\right)\right)^{2} - x_{1} x_{2} \left]\right.\)

=>\(\left(\right. x_{1} - x_{2} \left.\right) 2 \cdot \left[\right. 2^{2} - 2 \left(\right. m - 1 \left.\right) \left]\right. = \left(\right. x_{1} - x_{2} \left.\right) \left(\right. 2^{2} - \left(\right. m - 1 \left.\right) \left.\right)\)

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

=>\(\left(\right. x_{1} - x_{2} \left.\right) \left(\right. 12 - 4 m \left.\right) - \left(\right. x_{1} - x_{2} \left.\right) \left(\right. 5 - m \left.\right) = 0\)

=>\(\left(\right. x_{1} - x_{2} \left.\right) \left(\right. 12 - 4 m - 5 + m \left.\right) = 0\)

=>\(\left(\right. x_{1} - x_{2} \left.\right) \left(\right. - 3 m + 7 \left.\right) = 0\)

=>\(\left[\begin{array}{l}x_1-x_2=0\\ -3m+7=0\end{array}\right.\Leftrightarrow\left[\begin{array}{l}x_1=x_2\\ m=\frac{7}{3}\left(\right.loại\left.\right)\end{array}\right.\)

Nếu \(x_{1} = x_{2}\) thì \(x_{1} = x_{2} = \frac{x_{1} + x_{2}}{2} = \frac{2}{2} = 1\)

\(x_{1} x_{2} = m - 1\)

=>m-1=1

=>m=2(nhận)

Ta có: \(\Delta = \left(\left(\right. - 2 m \left.\right)\right)^{2} - 4 \cdot 1 \cdot \left(\right. 4 m - 4 \left.\right)\)

\(= 4 m^{2} - 16 m + 16 = \left(\left(\right. 2 m - 4 \left.\right)\right)^{2} > = 0 \forall m\)

=>Phương trình luôn có hai nghiệm

Theo Vi-et, ta có:

\(\left[\begin{array}{l}x_1+x_2=-\frac{b}{a}\\ x_1x_2=\frac{c}{a}=4m-4\end{array}\right.\)

\(x_{1}^{2} + x_{2}^{2} - 8 = 0\)

=>\(\left(\left(\right. x_{1} + x_{2} \left.\right)\right)^{2} - 2 x_{1} x_{2} - 8 = 0\)

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

=>\(4 m^{2} - 8 m = 0\)

=>4m(m-2)=0

=>\(\left[\begin{array}{l}m=0\\ m-2=0\end{array}\right.\Leftrightarrow\left[\begin{array}{l}m=0\\ m=2\end{array}\right.\)

Δ=\(b^2-4ac\) =[−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 \(x_{1} = m ; x_{2} = m + 2\)

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

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

=>\(\left[\begin{array}{l}3m+6=m\\ 3m+6=-m\end{array}\right.\)

=>\(\left[\begin{array}{l}2m=-6\\ 4m=-6\end{array}\right.\)

=>\(\left[\begin{array}{l}m=-3\\ m=-\frac32\end{array}\right.\)

Δ=\(b^2\) -4ac=\(\left(-m\right)^2\) −4(m−2)

\(=m^2-4m+8\)

\(=\left(m-2\right)^2+4\ge4>0\forall m\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

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

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

=>\(\left.\left(\right.x_1-x_2\right)^2=\left(2\sqrt{5}\right)^2\)

=>\(\left.\left(\right.x_1+x_2\right)^2-4x_1x_2=20\)

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

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

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

=>\(\left[\begin{array}{l}m-6=0\\ m+2=0\end{array}\right.\)

\(\left[\begin{array}{l}m=6\\ m=-2\end{array}\right.\)

Δ=\(b^2\) -4ac=\(\left(-m\right)^2\) −4(m−2)

\(=m^2-4m+8\)

\(=\left(m-2\right)^2+4\ge4>0\forall m\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

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

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

=>\(\left.\left(\right.x_1-x_2\right)^2=\left(2\sqrt{5}\right)^2\)

=>\(\left.\left(\right.x_1+x_2\right)^2-4x_1x_2=20\)

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

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

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

=>\(\left[\begin{array}{l}m-6=0\\ m+2=0\end{array}\right.\)

\(\left[\begin{array}{l}m=6\\ m=-2\end{array}\right.\)

Δ=\(b^2\) -4ac=\(\left(-m\right)^2\) −4(m−2)

\(=m^2-4m+8\)

\(=\left(m-2\right)^2+4\ge4>0\forall m\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

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

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

=>\(\left.\left(\right.x_1-x_2\right)^2=\left(2\sqrt{5}\right)^2\)

=>\(\left.\left(\right.x_1+x_2\right)^2-4x_1x_2=20\)

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

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

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

=>\(\left[\begin{array}{l}m-6=0\\ m+2=0\end{array}\right.\)

\(\left[\begin{array}{l}m=6\\ m=-2\end{array}\right.\)

Δ=b\(^2\) -4ac=\(\left(4\right)^2\) −4.2.m

=16-8m

=16-8m

Để phương trình có hai nghiệm thì Δ\(\ge\) 0

16-8m\(\ge\)0

-8m\(\ge\)-16

m\(\le\) 2

Theo Vi-et, ta có:

\({\begin{cases}x_1+x_2=-\frac{b}{a}=\frac{-4}{2}=-2\\ x_1x_2=\frac{c}{a}=m\end{cases}}\)

\(x_1^2+x_2^2=10\)

=>\(\left.\left(\right.x_1+x_2\right)^2-2x_1x_2=14\)

=>\(\left(-2\right)^2-2m=10\)

=>4-2m=10

=> m=-3(thoả mãn)