Question

（x-1)(x-2)(x-3)(x-4)-24 (换元法）

Answer 1

（x-1)(x-2)(x-3)(x-4)-24 (换元法）

第一题：（x-1）(x-4)(x-2)(x-3)=(x^2-5x+4)(x^2-5x+6)
设y=x^2-5x,则原式=(y+4)(y+6)-24=y^2+10y=y(y+10)
把y带入，则原式=（x^2-5x）（x^2-5x+10）=x(x-5)（x^2-5x+10）
第二题：原式=x^3-2x^2+X^2-4=x^2(x-2)+(x-2)(x+2)=(x-2)(x^2+x+2)

1∕(x-1)(x-2)+1∕(x-2)(x-3)+1∕(x-3)(x-4)

1∕(x-1)(x-2)+1∕(x-2)(x-3)+1∕(x-3)(x-4)
=[(x-3)(x-4)+(x-1)(x-4)+(x-1)(x-2)]/(x-1)(x-2)(x-3)(x-4)
=(x^2-7x+12+x^2-5x+4+x^2-3x+2)/(x-1)(x-2)(x-3)(x-4)
=(3x^2-15x+18)/(x-1)(x-2)(x-3)(x-4)
=3(x^2-5x+6)/(x-1)(x-2)(x-3)(x-4)
=3(x-2)(x-3)/(x-1)(x-2)(x-3)(x-4)
=3/(x-1)(x-4)

(x-1)(x-2)+(x-2)(x-3)-2(x-3)(x-4)<-4

前一个展开
x^2-3*x+2+(x^2-5*x+6)-2（x^2-7*x+12)<-4
6*x-16<-4
6*x<12
x<2
后一个展开
x^2-15*x+54-(x^2-8*x+7)>14*x-35
合并-7*x+47>14*x-35
-21*x>-82
x<82/21

(x-1)(x-2)(x-3)(x-4)-24 怎么解

(x-1)(x-2)(x-3)(x-4)-24
=[(x-1)(x-4)][(x-2)(x-3)]-24
=[(x^2-5x)+4][(x^2-5x)+6]-24
=(x^2-5x)^2+10(x^2-5x)+24-24
=(x^2-5x)^2+10(x^2-5x)
=(x^2-5x)(x^2-5x+10)
=x(x-5)(x^2-5x+10)

(x-1)(x-2)(x-3)(x-4)+1

原式=[(x-1)(x-4)][(x-2)(x-3)]+1
=[(x²-5x)+4][(x²-5x)+6]+1
=(x²-5x)²+10(x²-5x)+24+1
=(x²-5x)²+10(x²-5x)+25
=(x²-5x+5)²

(X-1)(X-2)(X-3)(X-4)-3

(X-1)(X-2)(X-3)(X-4)-3
=(X-1)(X-4)(X-2)(X-3)-3
=(X^2-5X+4)(X^2-5X+6)-3.
令t=X^2-5X,
原式=（t+4)(t+6)-3
=t^2+10t+21
=(t+3)(t+7)
=(X^2-5X+3)(X^2-5X+7).

换元法： (x-3)/(x+1)+(x+1)/(x-3)=2

设y=x-3/x+1变为y+1/y=2.等号两边同乘y解得y=x-3/x+1=1但x-3/x+1=1无解，所以这个题无解

(x(X-1)(X-2)(X-3)(X-4)+x(X-1)(X-2)) /x(X-1)(X-2)=43 求x

(X-3)(X-4)+1=43
x^2-7x-30=0
(X-10)(X+3)=0
所以 X= 10 或 —3

(x-1)(x-2)(x-3)(x-4)+1=0

(x-1)(x-2)(x-3)(x-4)+1
=(x*x-5x)平方+10(x*x-5x)+24+1
=(x*x-5x)平方+10(x*x-5x)+25
=(x*x-5x+5)平方
所以有:(X^2-5X+5)^2=0
X^2-5X+5=0
(X-5/2)^2=5/4
X1=5/2+根5/2
X2=5/2-根5/2

(x-1)(x-2)(X-3)(x-4)-48

（x-1）（x-2）（x-3）（x-4）-48
=（x-1）（x-4）（x-2）（x-3）-48
=（x^2-5x+4）（x^2-5x+6）-48
=（x^2-5x)^2+10（x^2-5x+6）-24
=（x^2-5x-2）（x^2-5x+12）