已知x-(1/x)=5,求x²+(1/x²)的值 已知(x+y)²=16,(x-y)²=4,求x²+xy+y²的值

问题描述:

已知x-(1/x)=5,求x²+(1/x²)的值 已知(x+y)²=16,(x-y)²=4,求x²+xy+y²的值

x2+(1\x)=27
x²+xy+y²=13

x²+(1/x²)
=(x-1/x)²+2
=25+2
=27
(x+y)²=x²+2xy+y²=16
(x-y)²=x²-2xy+y²=4
两式相加,得 x²+y²=10
两式相减,得 xy=3
x²+xy+y²
=10+3
=13