已知实数X,Y满足方程组X^3+Y^3=19,X+Y=1 则 X^2+Y^2=

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 11:22:28

已知实数X,Y满足方程组X^3+Y^3=19,X+Y=1 则 X^2+Y^2=
已知实数X,Y满足方程组X^3+Y^3=19,X+Y=1 则 X^2+Y^2=

已知实数X,Y满足方程组X^3+Y^3=19,X+Y=1 则 X^2+Y^2=
X^3+Y^3=19=(X+Y)(X^2+Y^2-XY),即X^2+Y^2-XY=19,两边同时乘2,
【1】2X^2+2Y^2-2XY=38,而X+Y=1=(X+Y)²=X^2+Y^2-2XY
【2】X^2+Y^2-2XY=1,【1】+【2】得出X^2+Y^2= 13

x³+y³=(x+y)(x²-xy+y²)=x²+y²-xy=x²+y²-[(x+y)²-x²-y²]/2=x²+y²-[1-x²-y²]/2=3(x²+y²)/2 -1/2=19
可以得出x²+y²=13

(x+y)(x²+y²)
=x³+xy²+x²y+y³
=x³+y³+xy(x+y)
=x³+y³+{[(x+y)²-(x²+y²)]/2}(x+y)
把条件代入得:
1×(x²+y²)=19+{[1²-(x²+y²)]/2}×1
解得x²+y²=13