已知一元二次方程ax平方＋bx＋c=0的两根之和为p，两根平方和为q,两根立方和为r,则ar＋bq＋cp的值是？
热心网友
解:设方程两跟为X1,X2,则p=X1+X2q=(X1)^2+(X2)^2r=(X1)^3+(X2)^3a(X1)^2+bX1+c=0 (1)a(X2)^2+bX2+c=0 (2)(1)*X1得a(X1)^3+b(X1)^2+cX1=0 (3)(1)*X1得a(X2)^3+b(X2)^2+cX2=0 (4)(3)+(4)得a[(X1)^3+(X2)^3]+b[(X1)^2+(X2)^2]+c(X1+X2)=0即ar＋bq＋cp=0
热心网友
p=x1+x2=-b/aq=x1^2+x2^2=(x1+x2)^2-2x1x2=(b/a)^2-2c/ar=x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^2)=(x1+x2)^3-3x1x2(x1+x2) =-(b/a)^3-3(-b/a)c/a =-(b/a)^3+3bc/a^2ar+bq+cp=-(b^3)/(a^2)+3bc/a+(b^3)/(a^2)-2bc/a-bc/a=0