Question

解方程:1/(X-4)+4/(X-1)=2/(X-3)+3/(X-2)

Answer 1

解：∵(x+1)(x+2)(x+3)(x+4)=120
==>[(x+1)(x+4)][(x+2)(x+3)]=120
==>(x²+5x+4)(x²+5x+6)=120
==>[(x²+5x)+4][(x²+5x)+6]=120
==>(x²+5x)²+10(x²+5x)+24=120
==>(x²+5x)²+10(x²+5x)-96=0
==>[(x²+5x)-6][(x²+5x)+16]=0
(应用十字相乘法分解因式)
==>(x²+5x-6)(x²+5x+16)=0
==>(x-1)(x+6)(x²+5x+16)=0
(再次应用十字相乘法分解因式)
∴x-1=0，或x+6=0
（x²+5x+16>0）
==>x1=1，x2=-6
故原方程的解是x1=1，x2=-6。

Answer 2

1/(X-4)+4/(X-1)=2/(X-3)+3/(x-2)

1/(x-4)-2/(x-3)=3/(x-2)-4/(x-1)

[(x-3)-2(x-4)]/(x-4)(x-3)=[3(x-1)-4(x-2)]/(x-2)(x-1)

(5-x)/(x^2-7x+12)=(5-x)/(x^2-3x+2)

(1):5-x=0
x=5

(2):x^2-7x+12=x^2-3x+2
x=2.5

经检验：方程的解是：X1=5，X2=2。5

Answer 3

没有简便算法
只能一步一步通分了
答案是5和2.5

Answer 4

5

Answer 5

X＝5