Question

已知x=√3+1,y=√3-1,求代数式x²-y²/x²y+xy²的值

Answer 1

解；原式=x²-y²/xy(x+y)

=x²-y/x(x+y)

当x=√3+1,y=√3-1,时.

原式=（√3+1)²-(√3-1,)/(√3+1）2√3

=4+2√3-（4-2√3）/4√3

=4+2√3-√3/3+1/2

=4.5+5√3/3