x=1/(2+√3)
=(2-√3)/[(2-√3)(2+√3)]
=(2-√3)/(2²-√3²)
=(2-√3)/(4-3)
=2-√3
∵0﹤x﹤1
∴x-1﹤0
1/x=2+√3
原式=[ (x²-2x+1)/(1-x) ]- [√(x²-2x+1)]/(x²-x)
=[ (1-x)²/(1-x) ]-[√(x-1)²]/[x(x-1)]
=1-x -|x-1|/[x(x-1)]
=1-x+(x-1)/[x(x-1)]
=1-x+(1/x)
=1-(2-√3)+2+√3
=1-2+√3+2+√3
=1+2√3