已知x+x^(-1)(←x的-1次方)=3,求下列各式的值:(1)x^(1/2)+x^(-1/2);(2)x^(1/2)-x^(-1/2).

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已知x+x^(-1)(←x的-1次方)=3,求下列各式的值:(1)x^(1/2)+x^(-1/2);(2)x^(1/2)-x^(-1/2).
已知x+x^(-1)(←x的-1次方)=3,求下列各式的值:
(1)x^(1/2)+x^(-1/2);
(2)x^(1/2)-x^(-1/2).

已知x+x^(-1)(←x的-1次方)=3,求下列各式的值:(1)x^(1/2)+x^(-1/2);(2)x^(1/2)-x^(-1/2).
[x^(1/2)+x^(-1/2)]^2=x+x^(-1)+2=5
因此x^(1/2)+x^(-1/2)=根号5
[x^(1/2)-x^(-1/2)]^2=x+x^(-1)-2=1
因此x^(1/2)-x^(-1/2)=正负1


x+x^(-1)=3知:X必为正数
令Y=X^(1/2)+X^(-1/2)
则Y^2=[X^(1/2)+X^(-1/2)]^2
=X^(+1) + X^(-1)+2
=3+2=5
∴Y=X^(1/2)+X^(-1/2)=5^(1/2)
同理:
令Y=X^(1/2)-X^(-1/2)
则Y^2=[X^(1/2)-X^(-1/2)]^2
=X^(+1) - X^(-1)-2
=3-2=1
∴Y=X^(1/2)-X^(-1/2)=1

分析:已知x+x^(-1)=3,可化为:
x+1/x=3,同时乘以x,得:
x^2-3x+1=0
解出x,然后把x代入(1),(2)就可以.
(b^2-4ac=(-3)^2-4*1*1=9-4=5>0,因此有两解.