已知y小于根号下x-4+根号下4-x+1化简y-1分之根号下y的平方-2y+1+1-y的绝对值+y的值
问题描述：
已知y小于根号下x-4+根号下4-x+1化简y-1分之根号下y的平方-2y+1+1-y的绝对值+y的值
答
根号下则x-4>=0,x.=4
4-x>=0,x,=4
所以x=4
所以yy-11-y>0
所以原式=√(y-1)²/(y-1)+|1-y|+y
=|y-1|/(y-1)+1-y+y
=-(y-1)/(y-1)+1
=-1+1
=0
答
∵y∴x-4≥0,4-x≥0
∴x-4=0,x=4
∴y∴y-1∴√(y²-2y+1)/(y-1)+|1-y|+y
=√(y-1)²/(y-1)+|y-1|+y
=|y-1|/(y-1)+|y-1|+y
=-(y-1)/(y-1)-(y-1)+y
=-1-y+1+y
=0