实数1/a,1,1/c成等差数列,实数a^2,1,c^2成等比数列,求(a+c)/(a^2+c^2)
人气:441 ℃ 时间:2019-09-29 02:36:44
解答
1/a+1/c=2 => a+c=2ac (a+c)^2=4a^2c^2=4 ==> a^2+c^2=4-2ac
(a+c)/(a^2+c^2)=2ac/[4-2ac]=ac/(2-ac)
a^2*c^2=1 ==> ac=1 ac=-1
(a+c)/(a^2+c^2)=ac/(2-ac) =1
推荐
- 实数1/a,a,1/c成等差数列,实数a^2,1,c^2成等比数列,则(a+c)/(a^2+c^2)=?
- 实数1/a,1,1/c成等差数列,实数a^2,1c^2成等比数列,则(a+c)/(a^2+c^2)=
- 三个不同的实数a、b、c成等差数列,且a、c、b成等比数列,求a:b:c.
- 实数1/a,1,1/c成等差数列实a^2,1,c^2成等比数列则(a+c)/(a^2 +c^2)=
- 已知实数a,b,c成等差数列,a+1,b+1,c+4成等比数列,且a+b+c=15,求a,b,c
- he was ___ (too much,much too) worried about his son.请问此处填啥?
- 一元二次方程kx2-(2k-1)x+k+2=0,当k为何值时,方程有两个不相等的实数根?
- 怎样让在地震中的房屋不容易倒塌
猜你喜欢