一元四次方程求解

X范围1<X<4

1/(4-X)(6-X)=X^2/(5-X)^2=（X^2-1）/1

根据相似三角形列出的方程，任意两个等式都应该可以求出结果

等候大家的解答（最好有过程）

X^2/(5-X)^2=（X^2-1）/1

X^2=(5-X)^2(X^2-1)

X^2=(X^2-10X+25)(X^2-1)

X^2=X^4-10X^3+25X^2-X^2+10X-25

X^4-10X^3+23X^2+10X-25=0

1/(4-X)(6-X)=X^2/(5-X)^2

X^2(4-X)(6-X)=(5-X)^2

X^2(X^2-10X+24)=X^2-10X+25

X^4-10X^3+24X^2=X^2-10X+25

X^4-10X^3+23X^2+10X=25

X^4-10X^3+25X^2-2X^2+10X=25

X^2(X-5)^2-2X(X-5)+1=26

[X(X-5)-1]^2=26

X(X-5)-1=√26

X^2-5X