一道初三数学题哦,帮帮忙啊

2024-11-30 23:53:57
推荐回答(4个)
回答1:

1.只要b^2-4ac=2*(m+1)*2*(m+1)-4*1*(m^2+2)=0
=>4m^2+8m+4-4m^2-8=0
=>m=1/2

2.只要b^2-4ac=2*(m+1)*2*(m+1)-4*1*(m^2+2)>0
=>4m^2+8m+4-4m^2-8>0
=>m>1/2就行了
例如m=1
=>原方程为x^2-4x+3=0
=>x1=1,x2=3

回答2:

要使有两个相等实根,b平凡-4ac=0
2(m+1)*2(m+1)-4*(m*m+2)=0
m=-1/2
(2)
2m+3=一个整数的平方
m=-1
x1=1,x2=-1

回答3:

1、△=b²-4ac=8m-4=0
m=0.5
2、△>0,则m>0.5,取m=1,
方程为 x²-4x+3=0 ,x1=1,x2=3

回答4:

(1)m=0.5,方程有两个相等的实数根
(2)m=1时,方程有两个不相等的实数根,分别是 x1=1、x2=3

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