x^2-2(m+1)x+m^2-2m-3=0两不等实根有一根为0,是否存在x^2-(k-m)x-k-m^2+5m-2=0两根XX1,X2且绝对值X1-X2=0
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![x^2-2(m+1)x+m^2-2m-3=0两不等实根有一根为0,是否存在x^2-(k-m)x-k-m^2+5m-2=0两根XX1,X2且绝对值X1-X2=0](/uploads/image/z/8892081-9-1.jpg?t=x%5E2-2%28m%2B1%29x%2Bm%5E2-2m-3%3D0%E4%B8%A4%E4%B8%8D%E7%AD%89%E5%AE%9E%E6%A0%B9%E6%9C%89%E4%B8%80%E6%A0%B9%E4%B8%BA0%2C%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8x%5E2-%28k-m%29x-k-m%5E2%2B5m-2%3D0%E4%B8%A4%E6%A0%B9XX1%2CX2%E4%B8%94%E7%BB%9D%E5%AF%B9%E5%80%BCX1-X2%3D0)
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x^2-2(m+1)x+m^2-2m-3=0两不等实根有一根为0,是否存在x^2-(k-m)x-k-m^2+5m-2=0两根XX1,X2且绝对值X1-X2=0
x^2-2(m+1)x+m^2-2m-3=0两不等实根有一根为0,
是否存在x^2-(k-m)x-k-m^2+5m-2=0两根X
X1,X2且绝对值X1-X2=0
x^2-2(m+1)x+m^2-2m-3=0两不等实根有一根为0,是否存在x^2-(k-m)x-k-m^2+5m-2=0两根XX1,X2且绝对值X1-X2=0
有两个不等的实根且一个为0.得C=0,b≠0.得(m+1)(m-3)=0且m+1≠0 m=3.
2)x^2-(k-3)X-K+4=0 Δ=0 (k-3)^2+4k-16=0其式有解
m x平方-2x-m+1
因式分解(m-1)x-(x^2-m)
x^2-(2m+1)x+m^2+m
若多项式5x*x-2mxy-3y*y+4xy-3x+1中不含xy项,求(-(m*m*m)+2(m*m)-m+1)-(m*m*m+2m*m-m+4)
|x-2|-|x-m|
计算x^3m/(x^m-1)-x^2m/(x^m+1)-1/(x^m-1)+1(x^m+1)
2x-3m
x^(3)*x^(2m +1)*x^(m)=x^(31) 求m的值
(1-m)x^2+(m-3)x+2 十字相乘
X2-(2m+1)x+2m
mx^2-(m+1)x-m
已知函数y=(m+1)x^m^2-3m-2+(m-1)x(m是常数)
计算:(1).(2x-3y)/(y-x)-(x+2y)/(x-y) (2).[5m/(m-3)+2m/(m+3)]*(m^2-9)/m
解关于x的不等式:x^2-(3m+1)x+2m^2+m>0
分解因式x^m+3-2x^m+2y+x^m+1y^2
分解因式:x^m+3-2x^m+2*y+x^m+1*y^2
分解因式x^m+3-(2x^m+2)y+(x^m+1)y^2
因式分解:mx^2-(m^2+m+1)x+1+m