已知m^2+2n+2p^2-2mn-2np-6p+9=0,求【（n^2）p】/m的值
问题描述：
已知m^2+2n+2p^2-2mn-2np-6p+9=0,求【（n^2）p】/m的值
答
sina=-1/2
又 sin²a+cos²a=1
从而 cosa=±√3/2
∴原式=1/cos(a-7π)=1/(-cosa)=±2√3/3ֻ�ǻ�����ֵ����m^2+2n+2p^2-2mn-2np-6p+9=0����Ȼ�������ã� �⣺�� m^2+2n^2+2p^2-2mn-2np-6p+9=0�� m^2-2mn+n^2+n^2+2p^2-2np-6p+9=0 (m-n)^2+n^2-2np+p^2+p^2-6p+9=0 (m-n)^2++(n-p)^2+(p-3)^2=0�Ӷ� m-n=0 n-p=0 p-3=0�� m=n=p=3�ࡾ��n^2��p��/m��ֵ =����p^2��p��/p =p^2 =3^2 =9