y=x^2+px+q经过两点M(m,m),N(n,n)(m不等于n):求m+n=1-p,mn=q;若点M是抛物线的顶点,
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/08 05:29:27
y=x^2+px+q经过两点M(m,m),N(n,n)(m不等于n):求m+n=1-p,mn=q;若点M是抛物线的顶点,n=3,求p,q
![y=x^2+px+q经过两点M(m,m),N(n,n)(m不等于n):求m+n=1-p,mn=q;若点M是抛物线的顶点,](/uploads/image/z/16772198-14-8.jpg?t=y%3Dx%5E2%2Bpx%2Bq%E7%BB%8F%E8%BF%87%E4%B8%A4%E7%82%B9M%28m%2Cm%29%2CN%28n%2Cn%29%EF%BC%88m%E4%B8%8D%E7%AD%89%E4%BA%8En%EF%BC%89%EF%BC%9A%E6%B1%82m%2Bn%EF%BC%9D1%EF%BC%8Dp%2Cmn%EF%BC%9Dq%EF%BC%9B%E8%8B%A5%E7%82%B9M%E6%98%AF%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E9%A1%B6%E7%82%B9%2C)
1)将(m,m)(n,n)代入,
m^2+pm+q=m,(1)
n^2+pn+q=n,(2)
两式相减,
(m^2+pm+q)-(n^2+pn+q)=m-n,(1)
m^2-n^2+p(m-n)=m-n,
(m+n)(m-n)+p(m-n)=m-n
因为m,n不相等,
所以m+n+p=1,
即m+n=1-p,
(1)+(2),
m^2+pm+q+n^2+pn+q=m+n
m^2+n^2+p(m+n)+2q=m+n
(m+n)^2-2mn+p(m+n)+2q=m+n
(1-p)^2-2mn+p(1-p)+2q=1-p
1-2p+p^2-2mn+p-p^2+2q=1-p
所以mn=q
2)n=3代入到m+n=1-p,得,
m+3=1-p,
又点M是抛物线的顶点,所以x=-b/2a=-p/2=m,
解得,
m=2,p=-4,
x=2,p=-4代入抛物线,
2^2-8+q=2,
q=6
所以p=-4,q=6
m^2+pm+q=m,(1)
n^2+pn+q=n,(2)
两式相减,
(m^2+pm+q)-(n^2+pn+q)=m-n,(1)
m^2-n^2+p(m-n)=m-n,
(m+n)(m-n)+p(m-n)=m-n
因为m,n不相等,
所以m+n+p=1,
即m+n=1-p,
(1)+(2),
m^2+pm+q+n^2+pn+q=m+n
m^2+n^2+p(m+n)+2q=m+n
(m+n)^2-2mn+p(m+n)+2q=m+n
(1-p)^2-2mn+p(1-p)+2q=1-p
1-2p+p^2-2mn+p-p^2+2q=1-p
所以mn=q
2)n=3代入到m+n=1-p,得,
m+3=1-p,
又点M是抛物线的顶点,所以x=-b/2a=-p/2=m,
解得,
m=2,p=-4,
x=2,p=-4代入抛物线,
2^2-8+q=2,
q=6
所以p=-4,q=6
已知:m*m=n+2,n*n=m+2.求m*m*m-2mn+n*n*n(m不等于n).
已知抛物线与x轴交于A(m,0),b(n,0)两点,与y轴交于C(0,3),点P是抛物线的顶点,若m-n=2,mn=3
已知:m,n是两个连续自然数(m<n),且q=mn.设p=q+n+q−m
已知m,n是两个连续自然数(m<n),且q=mn.设p=q+n+q−m
已知两点M(m,1/m),和N(n,1/n)(m不等于n)关于直线y=2x+b对称,求b的取值范围.
(m+n)(p+q)-(m+n)(p-q)=
m/x - n/(x+1)=0(m不等于n,mn不等于0)求x
若m,n是互为相反数,X,Y互为倒数,且n不等于0,|Z|=3,求XY(m+n)-m分之n+2XYZ的值
a(m-n)^p*(n-m)^q*(m-n)^q*(n-m)^p等于什么?
m,n,p,q都是实数,而且p×q=2(m+n).求证:x²+px+m=0,x²+qx+n=0.求证
2x+n/m=m/n-(m/n+n/m)x(m+n不等于0)
已知,m+1/(m+1)=n+ 1/(n-1)-2,且m-n+2不等于0,试求mn-m+n的值.