Question

解分式方程1/(x+2)+1/(2x+3)+1/(3x-4)=1/(6x+1)

立即
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Answer 1

1/(x+2)+1/(2x+3)=1/(6x+1)-1/(3x-4)2x+3+x+2/(x+2)(2x+3)=3x-4-6x-1/(6x+1)(3x-4)3x+5/(x+2)(2x+3)=-(3x+5)/(6x+1)(3x-4)(当3x+5=0时等式也成立所以x=-=5/3，(x+2)(2x+3)=-(6x+1)(3x-4)2x^2+7x+6=-(18x^2-21x-4)20x^2-14x+2=0,10x^2-7x+1=0(5x-1)(x-1)=0x=1/5和x=1/2或x=-5/3

Answer 2

题目稍作改正
1/6x+1
1/(x+2)+1/(2x+3)+1/(3x-4)=1/(6x+1)
1/(2x+3)+1/(3x-4)=1/(6x+1)
-1/(x+2)
5x-1/(2x+3)(3x-4)=
1-5x/(6x+1)(x+2)
(2x+3)(3x-4)=-(6x+1)(x+2)
6x²+7x+5=0
(2x-1)(3x+5)=0
x=1/2
x=
-5/3
检验
当
x=1/2
或
x=
-5/3
时
(x+2)(2x+3)(3x-4)(6x+1)≠0

Answer 3

2(x^2
+1/x^2)-3(x+
1/x)=1
2(x
+1/x)^2-3(x+
1/x)-5=0
[2(x
+1/x)-5](x+
1/x+1)=0
(1)
2x+
2/x -5=0
2x^2
-5x
+2=0
(2x-1)(x-2)=0
x1=1/2
x2=2
(2)
x+
1/x
+1=0
x^2
+x
+1=0
无解