maths 一題(10分)

( a - 1 )( a - 4 )x = a - 2 ( x + 1 )

( a - 1 )( a - 4 )x = a - 2x - 2

( a - 1 )( a - 4 )x + 2x = a - 2

( a^2 - 5a + 4 + 2 )x = a - 2

( a^2 - 5a + 6 ) x = a - 2

( a - 2 )( a - 3 )x = ( a - 2 )

x = ( a - 2 ) / ( a - 2 )( a - 3 )

x = 1 / ( a - 3 )


2007-09-05 19:39:38 補充：
1 restriction: a ≠ 3 as 1 / 0 is undefined.

2007-09-05 19:59:09 補充：
如果你是要以x表示a的話,( a - 1 )( a - 4 )x = a - 2 ( x + 1 )( a^2 - 5a + 4 )x = a - 2x - 2 xa^2 + ( - 5x - 1 )a + ( 6x + 2 ) = 0 用二次公式, a = { ( 5x + 1 )±√(( - 5x - 1 )^2 - 4 ( x )( 6x + 2 )) } / 2xa = {( 5x + 1 )±( 2x + 1 )} / 2x a = ( 7x + 2 ) / 2x 或 3 / 2

(a-1)(a-4)x=a-2(x+1)
a2次- 4a-a+5=a-2x-1
a2次-5a+5=a-2x-1
a2次-5a-a+5=-2x-1
a2次-6a+5=-2x-1
2x=-1-a2次+6a-5
x= (-1-a2次+6a-5)
x= (-6-a2次+6a) /2 （將它改成分數, means 2 is at the bottom)

a-1=0 x=a-4 0=0=0 a-4=-3 -3+x+1=-2 =-2

(a-1)(a-4)x=a-2(x+1)
(a^2-5a+4)x-a+2(x+1)=0
xa^2-(5x+1)a+6x+2=0
a=[(5x+1)+-(x+1)]/x-------by quadratic formula
a=4or 6+2/x

(a - 1)(a - 4)x = a - 2(x + 1)
(a - 1)(a - 4)x = a - 2x - 2
(a^2 - 5a + 4)x = a - 2x - 2
(a^2 - 5a + 6)x = a - 2
(a - 3)(a - 2)x = a - 2
a = 2, x can be any value,
a = 3 , eqt. has no real solution
a not equal to 2 and 3, x = 1/(a - 3)