已知2a^2+a-4=0, a-b=2, 求(1尀a+1)+(2尀b)的值

麻烦
2025-06-22 22:16:45
推荐回答(3个)
回答1:

xy^2/(x+y)的值为质数

令 x=ky
xy^2/(x+y)
=ky^2/(k+1)
令 y=p(k+1)

则 kp^2*(k+1)为质数
p=1, k=1
y=2
x=2
那么x+y仅有1种可能的值

x+y=4

回答2:

已知2a²+a-4=0
所以,-a=2a²-4
已知a-b=2,所以b=a-2
原式=1/(a+1)+2/(a-2)=[(a-2)+2(a+1)]/(a+1)(a-2)
=3a/(a²-a-2)
=3a/[a²+(2a²-4)-2]
=3a/(3a²-6)
=a/(a²-2)
=-(2a²-4)/(a²-2)
=-2

回答3:

-2
具体步骤见图