已知关于X的方程X^2-KX+K^2+N=0有两个不相等的实数根X1 X2 且（2X1+X2）^2-8(2X1+X2)+15=0 1.求证：N小于0 2.试用含K的代数式表示X1 3.当N=-3时,求K的值

已知关于X的方程X^2-KX+K^2+N=0有两个不相等的实数根X1 X2 且（2X1+X2）^2-8(2X1+X2)+15=0 1.求证：N小于0 2.试用含K的代数式表示X1 3.当N=-3时,求K的值
已知关于X的方程X^2-KX+K^2+N=0有两个不相等的实数根X1 X2 且（2X1+X2）^2-8(2X1+X2)+15=0 1.求证：N小于0 2.试用含K的代数式表示X1 3.当N=-3时,求K的值

已知关于X的方程X^2-KX+K^2+N=0有两个不相等的实数根X1 X2 且（2X1+X2）^2-8(2X1+X2)+15=0 1.求证：N小于0 2.试用含K的代数式表示X1 3.当N=-3时,求K的值
1.方程X^2-KX+K^2+N=0有两个不相等实根 所以(-K)^2-4*(K^2+N)>0 化简可得N

这个题的顺序不对，先给你解（2）问，再解（1）
（2） 设2x1+x2=t,
方程（2X1+X2）^2-8(2X1+X2)+15=0可化为 t^2-8t+15=0
解得 2x1+x2=t=5或3 ①
∵由韦达定理 x1+x2=k ② ∴①-②：x1=5-k或3-k
（1） 由（2）：2②-①：...

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这个题的顺序不对，先给你解（2）问，再解（1）
（2） 设2x1+x2=t,
方程（2X1+X2）^2-8(2X1+X2)+15=0可化为 t^2-8t+15=0
解得 2x1+x2=t=5或3 ①
∵由韦达定理 x1+x2=k ② ∴①-②：x1=5-k或3-k
（1） 由（2）：2②-①：x2=2k-5或2k-3
取x1=5-k x2=2k-5（都是正的，不影响结果，也可取t=3）
x1*x2=k^2+N=(5-k)(2k-5)③
③方程可化为 -（k-5/2）^2=(4N+25)/12
∴(4N+25)/12≤0 ∴4N＜0 ∴N＜0
（3）由上 N=-3代入③ k^2-3=(5-k)(2k-5)
解得 k=
(本人有点懒 最后结果 嘿嘿~~)

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证明：（1）∵关于x的方程x2-kx+k2+n=0有两个不相等的实数根，
∴△=k2-4（k2+n）=-3k2-4n＞0，
∴n＜- 3/4 k2
又-k2≤0，
∴n＜0．
（2）∵（2x1+x2）2-8（2x1+x2）+15=0，x1+x2=k，
∴（x1+x1+x2）2-8（x1+x1+x2）+15=0
∴（x1+k）2-8（x1+k...

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证明：（1）∵关于x的方程x2-kx+k2+n=0有两个不相等的实数根，
∴△=k2-4（k2+n）=-3k2-4n＞0，
∴n＜- 3/4 k2
又-k2≤0，
∴n＜0．
（2）∵（2x1+x2）2-8（2x1+x2）+15=0，x1+x2=k，
∴（x1+x1+x2）2-8（x1+x1+x2）+15=0
∴（x1+k）2-8（x1+k）+15=0
∴[（x1+k）-3][（x1+k）-5]=0
∴x1+k=3或x1+k=5，
∴x1=3-k或x1=5-k．
（3）∵n＜-3/4k2，n=-3，
∴k2＜4，即：-2＜k＜2．
原方程化为：x2-kx+k2-3=0，
把x1=3-k代入，得到k2-3k+2=0，
解得k1=1，k2=2（不合题意），
把x1=5-k代入，得到3k2-15k+22=0，△=-39＜0，所以此时k不存在．
∴k=1．

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