已知x=2y,y＝0，化简，根号y/(根号x-根号y)-根号y/(根号x+根号y)
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楼主、您好：
原式=√y/(√2y-√y)-√y/(√2y+√y)
=√y/[√y(√2-1)]-√y/[√y(√2+1)]
=1/(√2-1)-1/(√2+1)
=(√2+1)/(√2+1)(√2-1)-(√2-1)/(√2+1)(√2-1)
=(√2+1)/(2-1)-(√2-1)/(2-1)
=√2+1-√2+1
=2
希望能够帮到您。望采纳
原式=√y/(√2y-√y)-√y/(√2y+√y)
=√y/[√y(√2-1)]-√y/[√y(√2+1)]
=1/(√2-1)-1/(√2+1)
=(√2+1)/(√2+1)(√2-1)-(√2-1)/(√2+1)(√2-1)
=(√2+1)/(2-1)-(√2-1)/(2-1)
=√2+1-√2+1
=2
希望能够帮到您。望采纳
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