已知x=1是方程xx−1+kx−1＝xx+1

解分式方程xx+1−x+1x＝kx 当k=______时，分式方程xx−1+kx−1−xx+1＝0 若关于x的方程2kx−1−xx 用换元法解方程：（xx−1 已知：3x=xx-x+1求（xxxx+xx+1）分之xx 已知x/(xx+x+1)=a,求xx/(xxxx+xx+1)的值 已知xx+x-1=0,求:xx+1/xx,xxxx+1/xxxx 已知函数f（x）=xx−1． 分式方程xx−1−1 已知x,y是实数,且适合方程(xx+xy-12)(xx+xy-12)+(xy-2yy-1)(xy-2yy-1)=0求x, 解方程1/(xx+x)+1/(xx+3x+2)+1/(xx+5x+6)+1/(xx+7x+12)+1/(xx+9x+20 解分式方程1−xx−2+2＝12−x，可知方程（　　）