Question

已知x+y=1，x^2+y^2=2，求x^6+y^6的值

Answer 1

x+y=1，x²+y²=2
则：(x+y)²=x²+2xy+y² ---> 1=2+2xy
得：xy=-1/2，
则：x²y²=1/4
x^6+y^6=(x²+y²)(x^4-x²y²+y^4)
x^6+y^6=(x²+y²)²-2x²y²
则：x^4-x²y²+y^4=(x²+y²)²-3x²y²=4-3/4=13/4
所以：x^6+y^6=(x²+y²)(x^4-x²y²+y^4)=13/2

Answer 2

（x+y）²=1²
x²+2xy+y²=1
2xy=1-2
xy=-1/2
x的6次方+y的6次方
=(x²)³+(y²)³
=（x²+y²）(x的4次方-x²y²+y的4次方）
=（x²+y²）[(x²+y²)²-3x²y²]
=2×[2²-3×(-1/2)²]
=2×[4-3/4]
=8-3/2
=8-1.5
=6.5

Answer 3

x,y可求