1/x2+11x-8+1/x2+2x-8+1/x2-13x-8=0
解方程:X2--1/8(X2+2X)+X2+2X/3(X2--1)=11
1/x2+11x-8+1/x2+2x-8+1/x2-13x-8=0
已知x2-3x-1=0求x3-3x2-11x+8的值
解方程:1/(x2+2x+10)+1/(x2+11x+10)+1/(x2-13x+10)=0
解方程1/x2+2x+10+1/x2+11x+10+1/x2-13x+10=0
用换元法解方程8(x2+2x)x2-1+3(x2-1)x2+2x=11时若设x2-1x2+2x=y,则可得到整式方程是(
x2-5x+1=0则x2+x2/1
已知x2+4x-1=0,求2x四次幂-4x2-8x+1的值
f(2x-1)=x2+8 求f(x)
2( x2 +1/x2) - 9( x + 1/x) + 14 =
x2+x+1=2/(x2+x)解分式方程
解方程(x2+x)(x2+x-2)=-1