两道初三数学题1,设X1,X2是方程X^2+X+K=0的两个实数根,若恰有X1^2+X1X2+X2^2=2K^2成立,求K的值.2,已知X,Y均为实数,且满足XY+X+Y=17,X^2Y+XY^2=66,求：代数式X^4+X^3Y+X^2Y^2+XY^3+Y^4的值
2019-05-07
两道初三数学题
1,设X1,X2是方程X^2+X+K=0的两个实数根,若恰有X1^2+X1X2+X2^2=2K^2成立,求K的值.
2,已知X,Y均为实数,且满足XY+X+Y=17,X^2Y+XY^2=66,求：代数式X^4+X^3Y+X^2Y^2+XY^3+Y^4的值
优质解答
1.X1+X2=-1,X1*X2=K
X1^2+X1X2+X2^2=(X1+X2)^2-X1*X2=1-K=2K^2
即（2K-1）*（K+1）=0,
k1=-1,k2=1/2
Δ≥0,k≤1/4
所以k=-1
2.XY+(X+Y)=17,X^2Y+XY^2=xy(x+y)=66,
可知：xy,x+y分别为6,11,
X^4+X^3Y+X^2Y^2+XY^3+Y^4
=x^3（x+y)+x^2y^2+(x+y)y^3
=11(x^3+y^3)+36
=11(x+y)(x^2-xy+y^2）+36
=121[(x+y)^2-3xy]+36
=121*103+36
=12499
1.X1+X2=-1,X1*X2=K
X1^2+X1X2+X2^2=(X1+X2)^2-X1*X2=1-K=2K^2
即（2K-1）*（K+1）=0,
k1=-1,k2=1/2
Δ≥0,k≤1/4
所以k=-1
2.XY+(X+Y)=17,X^2Y+XY^2=xy(x+y)=66,
可知：xy,x+y分别为6,11,
X^4+X^3Y+X^2Y^2+XY^3+Y^4
=x^3（x+y)+x^2y^2+(x+y)y^3
=11(x^3+y^3)+36
=11(x+y)(x^2-xy+y^2）+36
=121[(x+y)^2-3xy]+36
=121*103+36
=12499