解分式方程(x+2⼀x+1)-(x+4⼀x+3)=(x+6⼀x+5)-(x+8⼀x+7)

这一步看不懂1+1⼀(x+1)-[1+1⼀(x+3)]=1+1⼀(x+5)-[1+1⼀(x+7)]
2024年11月15日 20:52
有4个网友回答
网友(1):

(x+2)/(x+1)=(x+1)/(x+1)+1/(x+1)=1+1/(x+1)
同理:(x+4)/(x+3)=1+1/(x+3)
(x+6)/(x+5)=1+1/(x+5)
(x+8)/(x+7)=1+1/(x+7)
(x+2/x+1)-(x+4/x+3)=(x+6/x+5)-(x+8/x+7)
1+1/(x+1)-[1+1/(x+3)]=1+1/(x+5)-[1+1/(x+7)]
方程可整理为:
(x+1)(x+3)=(x+5)(x+7) 且(x不等于-1,-3,-5,-7)
x²+4x+3=x²+12x+35
4x-12x=35-3
-8x=32
x=-4

网友(2):

方程所得集合为:{X|X ∈ R 且 X ≠ 0} , R为一切实数

网友(3):

网友(4):

解:
[(x+1)+1]/(x+1)-[(x+3)+1]/(x+3)=[(x+5)+1]/(x+5)-[(x+7)+1]/(x+7)
(x+1)/(x+1)+1/(x+1)-(x+3)/(x+3)-1/(x+3)=(x+5)/(x+5)+1/(x+5)-(x+7)/(x+7)-1/(x+7)
1+1/(x+1)-1-1/(x+3)=1+1/(x+5)-1-1/(x+7)
1/(x+1)-1/(x+3)=1/(x+5)-1/(x+7)
[(x+3)-(x+1)]/[(x+1)(x+3)]=[(x+7)-(x+5)]/[(x+5)(x+7)]
2/[(x+1)(x+3)]=2/[(x+5)(x+7)]
(x+1)(x+3)=(x+5)(x+7)
x²+4x+3=x²+12x+35
4x-12x=35-3
-8x=32
x=-4
经检验x=-4是原方程的根