一元二次方程.有过程最好.1.已知x1,x2是方程x2+kx+p=0的两根,x3,x4是方程x2+kx+q=0的两根,则(x1-x3)(x2-x4)(x1-x4)(x2-x3)=( )A.(p+q)2 B.p2+q2 C.(q-p)2 D.q2-p22.方程-m^4+4m2+2^nm2+2^n+5=0的正整数解有( )组.A.1 B,2 C.3 D.
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![一元二次方程.有过程最好.1.已知x1,x2是方程x2+kx+p=0的两根,x3,x4是方程x2+kx+q=0的两根,则(x1-x3)(x2-x4)(x1-x4)(x2-x3)=( )A.(p+q)2 B.p2+q2 C.(q-p)2 D.q2-p22.方程-m^4+4m2+2^nm2+2^n+5=0的正整数解有( )组.A.1 B,2 C.3 D.](/uploads/image/z/10764471-39-1.jpg?t=%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8B.%E6%9C%89%E8%BF%87%E7%A8%8B%E6%9C%80%E5%A5%BD.1.%E5%B7%B2%E7%9F%A5x1%2Cx2%E6%98%AF%E6%96%B9%E7%A8%8Bx2%2Bkx%2Bp%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%2Cx3%EF%BC%8Cx4%E6%98%AF%E6%96%B9%E7%A8%8Bx2%2Bkx%2Bq%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%2C%E5%88%99%28x1-x3%29%28x2-x4%29%28x1-x4%29%28x2-x3%29%3D%28+%29A.%28p%2Bq%292+B.p2%2Bq2+C.%28q-p%292+D.q2-p22.%E6%96%B9%E7%A8%8B-m%5E4%2B4m2%2B2%5Enm2%2B2%5En%2B5%3D0%E7%9A%84%E6%AD%A3%E6%95%B4%E6%95%B0%E8%A7%A3%E6%9C%89%28+%29%E7%BB%84.A.1+B%2C2+C.3+D.)
一元二次方程.有过程最好.1.已知x1,x2是方程x2+kx+p=0的两根,x3,x4是方程x2+kx+q=0的两根,则(x1-x3)(x2-x4)(x1-x4)(x2-x3)=( )A.(p+q)2 B.p2+q2 C.(q-p)2 D.q2-p22.方程-m^4+4m2+2^nm2+2^n+5=0的正整数解有( )组.A.1 B,2 C.3 D.
一元二次方程.有过程最好.
1.已知x1,x2是方程x2+kx+p=0的两根,x3,x4是方程x2+kx+q=0的两根,则(x1-x3)(x2-x4)(x1-x4)(x2-x3)=( )
A.(p+q)2 B.p2+q2 C.(q-p)2 D.q2-p2
2.方程-m^4+4m2+2^nm2+2^n+5=0的正整数解有( )组.
A.1 B,2 C.3 D.无数
3.若对任何实数a,关于x的方程x2-2ax-a+2b=0都有实数根,则实数b的取值范围是______.
4.设k为常数,关于x的方程x2-2x+(3k2-9k/x2-2x-2k)=3-2k 有四个不同的实数根,求k的取值范围.
5.已知关于x的方程x2+√(a-2009) x+?(a-2061)=0有两个实数根.求所有满足条件的实数a的值.
一元二次方程.有过程最好.1.已知x1,x2是方程x2+kx+p=0的两根,x3,x4是方程x2+kx+q=0的两根,则(x1-x3)(x2-x4)(x1-x4)(x2-x3)=( )A.(p+q)2 B.p2+q2 C.(q-p)2 D.q2-p22.方程-m^4+4m2+2^nm2+2^n+5=0的正整数解有( )组.A.1 B,2 C.3 D.
(1)由题意得x1+x2=-k x1*x2=p x3+x4=-k x3*x4=q推出得x1+x2=x3+x4推出得x1-x3=x4-x2 x1-x4=x3-x2题目中式子等于(x1-x3)^2*(x1-x4)^2=[(x1-x3)*(x1-x4)]^2=[x1^2-(x3+x4)x1+x3x4]^2=[x1^2+kx1+q]^2令x1^2+kx1+q=A有因为x1是的X^2+kx+p=0解所以x1^2+Kx1+p=0推出得A=q-p所以答案为(q-p)^2 (2)不会(3)由题意得B^2-4AC>=0推出得a^2+a-2b>=0又因为对任意a都有解所以B^2+4AC
题在哪……
(1)x1+x2=-k=x3+x4 所以x1-x3=x4-x2 x1-x4=x3-x2 (x1-x3)(x2-x4)(x1-x4)(x2-x3)=( x4-x2)(x2-x4)(x3-x2)(x2-x3)=( x4-x2)^2(x3-x2)^2=(x2^2-(x3+x4)x2+x4x3)^2=(-kx2+p+kx2+q)^2=(p+q)^2 选A
ax2+bx+c=0
(4) 设:x^2-2x-2k=t
则:t+(3k^2-9k)/t=3-4k
t^2+(4k-3)t+(3k^2-9k)=0
(t-3k)(t-(k-3))=0
t1=3k,t2=k-3
3k≠k-3,k≠-3/2
t=3k时
x^2-2x-2k=3k
x^2-2x-5k=0
判别式△=4+20k>0,k>-1/5
t...
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(4) 设:x^2-2x-2k=t
则:t+(3k^2-9k)/t=3-4k
t^2+(4k-3)t+(3k^2-9k)=0
(t-3k)(t-(k-3))=0
t1=3k,t2=k-3
3k≠k-3,k≠-3/2
t=3k时
x^2-2x-2k=3k
x^2-2x-5k=0
判别式△=4+20k>0,k>-1/5
t=k-3时
x^2-2x-2k=k-3
x^2-2x-3k+3=0
判别式△=4+4(3k-3)=12k-8>0,k>2/3
所以,k的取值范围是,k>2/3
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