1+1÷1x2+2=?

若 x2-3x+1=0,求（2x2×x3-5x2×x2+2 x3-8x2 ）÷（x2+1 ）的值 y=x+√(1-x2) y2=x2+(1-x2)+2x√(1-x2) 解方程：X2--1/8(X2+2X)+X2+2X/3(X2--1)=11 已知x2－4x+1=0,求x2／(x2)2－3x2+1 解方程 2/(x2-x)+6/(1-x2)=7/(x2+x)x2为的平方 x2-3x-1=0,求①x2 1/x2;②x2-1/x2 y=3x/1+x2的值域1-X2/1+X2→2-(X2+1)/1+X2→-1 + 2/(1+X2)这一步怎么得到的, 1,x1,x2...Xn,成等比数列,x1 x2..xn>0,x1*x2*...xn=?x1,x2...Xn,2成等比数列,x1 x2..xn>0,x1*x2*...xn=? x2-3x+1=0 x2/(x4-x2+1) x2-5x+1=0则x2+x2/1 7/x2+x+1/x2-x=6/x2-1 1+1÷1x2+2=? 2/1x2/1x2/1=（ ）=（ ） 1+1x2-1x2-6=? 2( x2 +1/x2) - 9( x + 1/x) + 14 = y=(x2-2x+1)/( x2+1)的值域 y=(1-x2)/(1+x2)的值域是?我知道把y=(1-x2)/(1+x2)转换成y=-(1+x2)+2/(1+x2),但是不知道y=-(1+x2)+2/(1+x2)怎么解, 2x2+1+2+3+2x2=