x 4 的解是x 7 2 方程 1 x 4 的解是x

時間 2021-08-11 17:28:52

1樓:匿名使用者

(1)x=(a+d)/2

第一個方程中方程兩邊的分母中常數項的和都是7,x的值是7/2,而(1/x-a)-(1/x-b)=(1/x-c)-(1/x-d)可變形為:(1/x-a)+(1/x-d)=(1/x-c)+(1/x-b),其中a+d=b+c,所以x=(a+d)/2

(2)x=7/2

(x-1)/(x-2)-(x-3)/(x-4)=(x-2)/(x-3)-(x-4)(x-5)

移項得:(x-1)/(x-2)+(x-4)(x-5)=(x-2)/(x-3)+(x-3)/(x-4)

變形得:-(x-1)/(x-2)-(x-4)(x-5)=-(x-2)/(x-3)-(x-3)/(x-4)

方程兩邊都加2:1-(x-1)/(x-2)+1-(x-4)(x-5)=1-(x-2)/(x-3)+1-(x-3)/(x-4)

整理:1/(x-2)+1/(x-5)=1/(x-3)+1/(x-4)

x=7/2

請採納。

2樓:

解:1、

1/(x-2)+1/(x-5)=1/(x-3)+1/(x-4)

等號兩邊的分母常數項分別為2+5=7,3+4=7,解x=7/2

1/(x-7)-1/(x-5)=1/(x-6)-1/(x-4)變形後得1/(x-7)+1/(x-4)=1/(x-6)+1/(x-5)

等號兩邊的分母常數項分別為7+4=11,6+5=11,解x=7/2

根據以上兩個算式得以下規律:

當等號兩邊的算式均為兩項相加,分子為1(或分子為t時),分母為(x-常數),並且等號兩邊算式得分母常數項和m與n相等時,方程的解x=t*m/2或x=t*n/2

所以變形方程(1/x-a)-(1/x-b)=(1/x-c)-(1/x-d)得,1/(x-a)+1/(x-d)=1/(x-b)+1/(x-c)

根據以上推出的規律的以上方程得解

x=(a+d)/2或x=(b+c)/2

2、(x-1)/(x-2)-(x-3)/(x-4)=(x-2)/(x-3)-(x-4)/(x-5)

變形以上方程得

(x-1)/(x-2)+(x-4)/(x-5)=(x-2)/(x-3)+(x-3)/(x-4)

化簡方程過程為以下:

(x-2+1)/(x-2)+(x-5+1)/(x-5)=(x-3+1)/(x-3)+(x-4+1)/(x-4)

(x-2)/(x-2)+1/(x-2)+(x-5)/(x-5)+1/(x-5)=(x-3)/(x-3)+1/(x-3)+(x-4)/(x-4)+1/(x-4)

1+1/(x-2)+1+1/(x-5)=1+1/(x-3)+1+1/(x-4)

1/(x-2)+1/(x-5)=1/(x-3)+1/(x-4)

根據第一個問題得出的規律

方程得解

x=(2+5)/2=7/2

3樓:日出伏羲

(1)x=(a+d)/2=(b+c)/2

(2)x=7/2

解方程:1/(x-6)-1/(x-5)=1/(x-4)-1/(x-3)

4樓:雪之凝

解:bai1/(x-6)-1/(x-5)=1/(x-4)-1/(x-3)

(x-5-x+6)/(x-5)(x-6)=(x-3-x+4)/(x-3)(x-4)

1/(x-5)(x-6)=1/(x-3)(x-4)(x-5)(x-6)=(x-3)(x-4)x^2-7x+12=x^2-11x+30

解得du:x=9/2

希望能zhi幫上你

,不明dao白還可以專追問,呵屬呵

5樓:匿名使用者

解:1/(x-6)-1/(x-5)=1/(x-4)-1/(x-3)[(x-5)-(x-6)]/[(x-6)(x-5)]=[(x-3)-(x-4)]/[(x-4)(x-3)]

(x-5-x+6)/[(x-6)(x-5)]=(x-3-x+4)/[(x-4)(x-3)]

1/[(x-6)(x-5)]=1/[(x-4)(x-3)](x-6)(x-5)=(x-4)(x-3)x²-11x+30=x²-7x+12

-11x+7x=12-30

-4x=-18

x=18/4

x=9/2

6樓:丁槐邰翔

3/x+6/x-1=x+5/x(x-1)

(9x-3)/x(x-1)

=(x+5)/x(x-1)

9x-3

=x+58x=

8x=1經檢驗,x=1時,方程分母為0,無意義

所以該方程無解

(1)x-3/x-1+x²-2/(x-1)(x-7)=1+x-5/x-7 (2)2x-1/x+4-5=96/x²-16-1-3x/4-x

7樓:小百合

(1)(x-3)/(x-1)+(x²-2)/(x-1)(x-7)=1+(x-5)/(x-7)

兩邊同乘以(x-1)(x-7)得:

(x-3)(x-7)+x²-2=(x-1)(x-7)+(x-5)(x-1)

4x=-7

x=-7/4

代入原方程驗證:x=-7/4是原方程的解。

(2)(2x-1)/(x+4)-5=96/(x²-16)-(1-3x)/(4-x)

兩邊同乘以(x²-16)得:

(2x-1)(x-4)+5(x²-16)=96+(1-3x)(x+5)

2x²-5x-37=0

x=(5±√321)/4

代入原方程驗證:x=(5±√321)/4都是原方程的解。

解方程(x-1/x-2)+(x-7/x-8)=(x-3/x-4)+(x-5/x-6)

8樓:匿名使用者

(x-1)/(x-2)+(x-7)/(x-8)=(x-3)/(x-4)+(x-5)/(x-6)

(x-2+1)/(x-2)+(x-8+1)/(x-8)=(x-4+1)/(x-4)+(x-6+1)/(x-6) (此步對分子加1減1,是簡便演算法)

1+1/(x-2)+1+1/(x-8)=1+1/(x-4)+1+1/(x-6)

1/(x-2)+1/(x-8)=1/(x-4)+1/(x-6)

1/(x-8)-1/(x-4)=1/(x-6)-1/(x-2) (此步移項成左右相減,是再次簡便演算法)

(x-4-x+8)/(x-8)(x-4)=(x-2-x+6)/(x-6)(x-2)

4/(x-8)(x-4)=4/(x-6)(x-2)

(x-8)(x-4)=(x-6)(x-2)

x^2-12x+32=x^2-8x+12

4x=20

x=5驗算:

1/3-1/3=1-1(驗算也正確)

9樓:匿名使用者

(x-1)/(x-2)+(x-7)/(x-8)=(x-3)/(x-4)+(x-5)/(x-6)

1+1/(x-2)+1+1/(x-8)=1+1/(x-4)+1+1/(x-6)

註釋 (x-1)/(x-2)=[(x-2)+1]/(x-2)=(x-2)/(x-2)+1/(x-2)=1+1/(x-2)

1/(x-2)+1/(x-8)=1/(x-4)+1/(x-6)

1/(x-8) -1/(x-6) =1/(x-4)-1(x-2)

[(x-6) -(x-8)]/[(x-6)(x-8)]=[(x-2)-(x-4)]/[(x-4)(x-2)]

2/[(x-6)(x-8)]=2/[(x-4)(x-2)]

(x-6)(x-8)=(x-4)(x-2)

x²-14x+48=x²-6x+8

x=5檢驗:

方程(x-1/x-2)-(x-3/x-4)=(x-2/x-3)-(x-4/x-5)解為x=7/2,(1/x-7)-(1/x-5)=(1/x-6)-(1/x-4)解為x=11/2

10樓:匿名使用者

(1/x-7)+(1/x-1)=(1/x-6)+(1/x-2)1/(x-7)-1/(x-6)=1/(x-2)-1/(x-1)(x-6-x+7)/[(x-7)(x-6)]=(x-1-x+2)/[(x-1)(x-2)]

x方-3x+2=x方-13x+42

10x=40

x=4(1/x+a)-(1/x+b)=(1/x+c)-(1/x+d)(x+b-x-a)/[(x+a)(x+b)]=(x+d-x-c)/[(x+c)(x+d)]

b-a=d-c

x方+(a+b)x+ab=x方+(c+d)x+cdx=(cd-ab)/(a+b-c-d)

11樓:

1/(x-7)-1/(x-5)=1/(x-6)-1/(x-4)[(x-5)-(x-7)]/[(x-5)*(x-7)]=[(x-4)-(x-6)]/[(x-4)*(x-6)]

2/(x^2-12x+35)=2/(x^2-10x+24)x^2-12x+35=x^2-10x+24-2x+11=0

x=11/2

x 4 的解是x 7 2 方程 1 x 4 的解是x

1 是4 2 是 a b 2 1 x 4 2 x a d 2 c b 2 這題是找規律嗎?還是要證明 1 1 x 7 1 x 6 1 x 2 1 x 1 移項 通分 x 7 x 6 x 2 x 1 x 4 2 同樣是先移項,然後你會發現,只有當其中一個分式的分子 0是方程在有解,即x a d 2或 ...

方程(x 3)(x 1)6x 4的解是

風歸雲 x 2 4x 3 6x 4 x 2 2x 1 0 x 2 4 4 2 2 2 2 2 x1 1 2 x2 1 2 方程 x 3 x 1 6x 4的解是 x1 1 2,x2 1 2 2y 1 2 3 2y 1 2 0 2y 1 1 2y 1 2 02y 2y 1 0 y1 0 y2 1 2 方...

4X 10 5x 6解方程過程,4x 10 15解方程?

6 10 5x 4x 解得x 16 希望採納 5x 4x 10 6 x 16 5x 4x 10 6 x 16 4x 10 15解方程?4x 15 10 4x 5 x 5 4 4x 10 5x 2解方程 4x 5x 10 2 x 12 x 12 解 x 12 4x 16 6x 10解方程要過程 解 6...