Question

天才顶吖……kd/2=[kd+k(d+d')]/2*d'可以得出d’=（根号2-1）d

Answer 1

两边乘2得kd=2kd*d'+kd'*d',

两边除以k得d=2d*d'+d'*d',

变形得d'*d'+2d*d'-d=0,

用公式法解得d'=(根号2-1)d和d'=-(根号2+1)d.