解方程:1⼀(X-4)+4⼀(X-1)=2⼀(X-3)+3⼀(X-2)

2024-11-29 08:39:59
推荐回答(5个)
回答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。

回答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

回答3:

没有简便算法
只能一步一步通分了
答案是5和2.5

回答4:

5

回答5:

X=5