→ love15 :謝謝您! 04/24 23:21
※ 引述《love15 ( )》之銘言:
: 3、三角形ABC中 角C=90度 角平分線AD BE 相交於I
: 若三角形ABI面積=12 求四邊形ABDE面積為何?
: 4、已知實數a、b、c滿足a+b+c=1 a^2+b^2+c^2=1
: 則a^5+b^5+c^5=p+q‧abc 其中p、q均為整數 求數對(p,q)=?
2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)=0
then a,b,c are the roots of
x^3-x^2-abc=0
hence,
a^3-a^2=abc...........(1)
b^3-b^2=abc...........(2)
c^3-c^2=abc...........(3)
then
(a^3+b^3+c^3)-(a^2+b^2+c^2)=3abc
a^3+b^3+c^3=1+3abc.....(4)
(1)*a+(2)*b+(3)*c we have
(a^4+b^4+c^4)-(a^3+b^3+c^3)=(a+b+c)abc=abc
a^4+b^4+c^4=1+4abc.......(5)
(1)*a^2+(2)*b^2+(3)*c^2 we have
(a^5+b^5+c^5)-(a^4+b^4+c^4)=(a^2+b^2+c^2)abc=abc
a^5+b^5+c^5=1+5abc........(6)
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