若x=根号5/2,求根号x+1-根号x-1/根号x+1+根号x-1 + 根号x+1+根号x-1/根号x+1-根号x-1
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/11/12 02:21:50
若x=根号5/2,求根号x+1-根号x-1/根号x+1+根号x-1 + 根号x+1+根号x-1/根号x+1-根号x-1 的值
[√(x+1)-√(x-1)]/[√(x+1)+√(x-1)] + [√(x+1)+√(x-1)]/[√(x+1)-√(x-1)]
=[√(x+1)-√(x-1)]^2/[√(x+1)+√(x-1)][√(x+1)-√(x-1)]
+ [√(x+1)+√(x-1)]^2/[√(x+1)-√(x-1)][√(x+1)+√(x-1)]}
=[√(x+1)-√(x-1)]^2/2
+ [√(x+1)+√(x-1)]^2/2
=[x+1+x-1-2√(x+1)√(x-1)]/2
+[x+1+x-1+2√(x+1)√(x-1)]/2
=[2x-2√(x+1)√(x-1)]/2+[2x+2√(x+1)√(x-1)]/2
=x-√(x+1)√(x-1)+x-√(x+1)√(x-1)
=2x
=2*√5/2
=√5
=[√(x+1)-√(x-1)]^2/[√(x+1)+√(x-1)][√(x+1)-√(x-1)]
+ [√(x+1)+√(x-1)]^2/[√(x+1)-√(x-1)][√(x+1)+√(x-1)]}
=[√(x+1)-√(x-1)]^2/2
+ [√(x+1)+√(x-1)]^2/2
=[x+1+x-1-2√(x+1)√(x-1)]/2
+[x+1+x-1+2√(x+1)√(x-1)]/2
=[2x-2√(x+1)√(x-1)]/2+[2x+2√(x+1)√(x-1)]/2
=x-√(x+1)√(x-1)+x-√(x+1)√(x-1)
=2x
=2*√5/2
=√5
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