កំណែ​លំហាត់​សិស្ស​ពូកែ​កម្ពុជា​២០១២៖ ពហុធា មានឫស​ប្រាំ​ផ្សេង​គ្នា

Comments

1. Sokha says:
   April 12, 2012 at 4:19 am

   The other way to solve this problem

   We have r(x)=(x-a1)(x-a2)(x-a3)(x-a4)(x-a5) (1)
   And s(x)=-x^2+5
   So s(x)=(sqrt(5)-x)(sqrt(5)+x) (2)

   From (1),
   r(sqrt(5))=(sqrt(5)-a1)(sqrt(5)-a2)(sqrt(5)-a3)(sqrt(5)-a4)(sqrt(5)-a5) (3)
   r(-sqrt(5))=-(sqrt(5)+a1)(sqrt(5)+a2)(sqrt(5)+a3)(sqrt(5)+a4)(sqrt(5)+a5) (4)

   From (2),(3), & (4) => p= -[(3)*(4)]=-r(sqrt(5))*r(-sqrt(5))
   So calculate r(sqrt(5)) & r(-sqrt(5)) , then we can find p.

   Reply
   • វិចិត្រ says:
     April 12, 2012 at 9:33 am

     this is definitely the better solution that i must update to this

     Reply
2. Rainy says:
   April 12, 2012 at 11:04 am

   ដូច​បេះ​បិទ

   Reply
3. Rath says:
   April 17, 2012 at 4:25 am

   ខ្ញុំធ្វើអញ្ខឹងដែរ

   Reply