已知x1,x2(x1
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已知x1,x2(x1
已知X1、X2(X1〈X2)是二次方程X^2-(m-1)X+n=0③的两个实数根,Y1、Y2是方程Y^2-(n+1)Y-6m=0⑤的两个实数根
所以X1 + X2 =m-1,X1 * X2 = n ,Δ = (m - 1)^2 - 4n > 0
Y1 + Y2 =n+1,Y1 * Y2 = -6m
又因为X1-Y1=2 ①,Y2-X2=2②
①-②,得X1 - Y1 - Y2 + X2 = 0
m - 1 -(n + 1)= 0
m - n = 2
m = n + 2④
④代入③,得X^2 - (n + 2 -1)X + n = 0
X^2 - (n + 1)X + n = 0
因式分解,得(X - n)(X - 1) = 0
X1 = n,X2 = 1 或 X1 = 1,X2 = n
验增根:假如X2 = 1代入②,得Y2 = 3,
再把Y2 =3代入⑤,得9 - 3 * (n + 1)- 6m =0
3n + 6m = 6
n + 2m = 2
因m = n + 2,得n = -2/3,m = 4/3
因X1〈X2,所以n = -2/3,m = 4/3
假如X1 = 1代入①,得Y1 = -1,
再把Y1 =-1代入⑤,得1 + 1 * (n + 1)- 6m =0
n - 6m = -2
因m = n + 2,得n = -2,m = 0
因X1〈X2,所以n = -2,m = 0为增根
所以n = -2/3,m = 4/3
所以X1 + X2 =m-1,X1 * X2 = n ,Δ = (m - 1)^2 - 4n > 0
Y1 + Y2 =n+1,Y1 * Y2 = -6m
又因为X1-Y1=2 ①,Y2-X2=2②
①-②,得X1 - Y1 - Y2 + X2 = 0
m - 1 -(n + 1)= 0
m - n = 2
m = n + 2④
④代入③,得X^2 - (n + 2 -1)X + n = 0
X^2 - (n + 1)X + n = 0
因式分解,得(X - n)(X - 1) = 0
X1 = n,X2 = 1 或 X1 = 1,X2 = n
验增根:假如X2 = 1代入②,得Y2 = 3,
再把Y2 =3代入⑤,得9 - 3 * (n + 1)- 6m =0
3n + 6m = 6
n + 2m = 2
因m = n + 2,得n = -2/3,m = 4/3
因X1〈X2,所以n = -2/3,m = 4/3
假如X1 = 1代入①,得Y1 = -1,
再把Y1 =-1代入⑤,得1 + 1 * (n + 1)- 6m =0
n - 6m = -2
因m = n + 2,得n = -2,m = 0
因X1〈X2,所以n = -2,m = 0为增根
所以n = -2/3,m = 4/3
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