数学题目~~
设实数s,t分别满足19s^2+99s+1=0和t^2+99t+19=0,且st不等于1,求(st+4s+1)/t的值
(19s^2是19乘以s的2次,t^2是t的2次)
参考答案:s,t是19x^2+99x+19=0的根
因为st不等于1 所以s=t
s^2+99s+1=0
s^2+1=-99s
s+1/s=-99
(st+4s+1)/t
=(s^2+4s+1)/s
=s+1/s+4
=-95
设实数s,t分别满足19s^2+99s+1=0和t^2+99t+19=0,且st不等于1,求(st+4s+1)/t的值
(19s^2是19乘以s的2次,t^2是t的2次)
参考答案:s,t是19x^2+99x+19=0的根
因为st不等于1 所以s=t
s^2+99s+1=0
s^2+1=-99s
s+1/s=-99
(st+4s+1)/t
=(s^2+4s+1)/s
=s+1/s+4
=-95