高一数学题目,很急啊,在线等~!

2024-12-21 02:29:19
推荐回答(6个)
回答1:

设三边为a b c, a=5

b c为方程2x^2-12x+m=0的两个根
b+c = 6
b-c < a =5
(b-c)^2 = (b+c)^2 -4bc = 36 - 4*(m/2) < 25
m > 11/2
delta = 12*12 -4*2*m = 144 - 8m >=0
m <= 18

11/2 < m <= 18

回答2:

三角形两边之和大于第三边,两边之差小于第三边,并且边长均为正。

记f(x)=2x^2-12x+m,设f(x)=0的二根为x1,x2,则x1*x2>0,所以m>0.

x1+x2=6>5,0=<|x1-x2|=sqrt((x1+x2)^2-4x1*x2)=sqrt(36-2m)<5

可得11/2

回答3:

另外两边存在
所以Δ≥0
所以Δ=144-8m≥0 所以 m≤18
又因为 两边之和大于第三边 两边之差小于第三边
所以 x1+x2=6 x1*x2=m/2
两边之和 6>5
所以 两边之差 |x1-x2|²=6²-2m<25
所以 m>11/2
所以范围为 11/2<m≤18

回答4:

解 因为方程2x^2-12x+m=0有两个根,设为a,b,则a+b=6 ab=m/2
由均值定理得 ((a+b)/2)^2>=ab=m/2,m<=2(3^2)=18
所以 m<=18

回答5:

假设另外两边长为x1和x2,则由根与系数的关系知x1+x2=6,x1*x2=m/2.
又因为三角形两边之差小于第三边,故有
|x1-x2|<5
故(x1-x2)^2<25

(x1-x2)^2=(x1+x2)^2-4*x1x2=6^2-4*m/2=36-2*m<25
解之得
m>5.5
再由差别式知m<=18
综上知,5.5

回答6:

用M表示2个根,其范围是12-根号下(144-8M)大于0,且两根之差大于5,两根之和大于5也就是-b/a=6所以成立至于要上面等式成立即可,自己在练习本上划拉划拉吧!

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