已知x+y²+4x-6y+13=0,求立方根下x²+1/x²
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已知x+y²+4x-6y+13=0,求立方根下x²+1/x²
已知x+y²+4x-6y+13=0,求立方根下x²+1/x²
已知x+y²+4x-6y+13=0,求立方根下x²+1/x²
x+y²+4x-6y+13=0
x+4x+4+y²-6y+9=0
(x+2)²+(y-3)²=0
则有
x=-2
y=3
所以
立方根下x²+1/x²
=立方根下(4+1/4)
=立方根下(17/4)
x+y²+4x-6y+13=(x+2)^2+(y-3)^2=0,所以x=-2,y=3,故x^2+1/x^2=17/4
第一个x是x^2的话、那就化简一下(x+2)^2+(y-3)^2=0
即表示(-2,3)这一点
x=2、x²+1/x²=4.25
x^2+y^2+4x-6y+13=0
x^2+4x+4+y^2-6y+9=0
(x+2)^2+(y-3)^2=0
所以x=-2,y=3,则3√x^2+1/x^2=3√4+1/4=3√17/4=3√17*2/4*2=3√34/2