Question

3(x-4)=x+6

解下列方程：（1）( 3y+5/3y-6)=(1/2)+(5y-4/2y-4) (2)( 4/x^2-4)+(x+3/x-2)
(3) (3/x)+(6/x-1)-{x-4/x(x-1)}=0 (4) (1/x+4)-(1/x+7)=(1/x+3)-(1/x+6)
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Answer 1

（1）( 3y+5/3y-6)=(1/2)+(5y-4/2y-4)
(3y+5)/3(y-2)=1/2+(5y-4)/2(y-2)
2(3y+5)=3(y-2)+3(5y-4)
6y+10=3y-6+15y-12
6y-18y=-18-10
-12y=-28
y=7/3
(2)( 4/x^2-4)+(x+3/x-2)
这个题目不完全
(3) (3/x)+(6/x-1)-{x-4/x(x-1)}=0
3/x+6/(x-1)-(x-4)/x(x-1)=0
3(x-1)+6x-(x-4)=0
3x-3+6x-x+4=0
8x=-1
x=-1/8
(4) (1/x+4)-(1/x+7)=(1/x+3)-(1/x+6)
1/(x+4)+1/(x+6)=1/(x+3)+1/(x+7)
(x+4+x+6)/(x+4)(x+6)=(x+3+x+7)/(x+3)(x+7)
(2x+10)/(x²+10x+24)=(2x+10)/(x²+10+21)
x²+10+21=x²+10+24
∴此方程无解.