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
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![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](/uploads/image/z/1673221-13-1.jpg?t=%E5%B7%B2%E7%9F%A5x-%EF%BC%881%2Fx%EF%BC%89%3D5%2C%E6%B1%82x%26%23178%3B%2B%EF%BC%881%2Fx%26%23178%3B%EF%BC%89%E7%9A%84%E5%80%BC+%E5%B7%B2%E7%9F%A5%EF%BC%88x%2By%EF%BC%89%26%23178%3B%3D16%2C%EF%BC%88x-y%EF%BC%89%26%23178%3B%3D4%2C%E6%B1%82x%26%23178%3B%2Bxy%2By%26%23178%3B%E7%9A%84%E5%80%BC)
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
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
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/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
x2+(1\x)=27
x²+xy+y²=13