I don't know why I got mine wrong from the answer key.
I found the common denominators to be 2(x+1)(x-1), hence, I calculated my numerator to be 1-x.
So I should get -1/2(x+1)
But it's really off the answer.
[(x+2)/(x^2 -1)] - [3/(2x-2)] = ?
2009-08-18 2:50 pm
回答 (9)
2009-08-18 2:56 pm
✔ 最佳答案
[ (x + 2) / (x² - 1) ] - [ 3 / (2x - 2) ] =[ (x + 2) / ((x - 1) (x + 1)) ] - [ 3 / (2 (x - 1)) ] =
[ (2 (x + 2)) / (2 (x - 1) (x + 1)) ] - [ (3 (x + 1)) / (2 (x - 1) (x + 1)) ] =
[ 2 (x + 2) - 3 (x + 1) ] / [ 2 (x - 1) (x + 1) ] =
[ 2x + 4 - 3x - 3 ] / [ 2 (x - 1) (x + 1) ] =
[ - x + 1 ] / [ 2 (x - 1) (x + 1) ] =
[ - (x - 1) ] / [ 2 (x - 1) (x + 1) ] =
- 1 / [ 2 (x + 1) ]
*******
2009-08-18 10:05 pm
= ([x + 2]/[x² - 1]) - (3/[2x - 2])
= ([x + 2]/[{x + 1}{x - 1}]) - (3/[2{x - 1})
= (2x + 4 - 3x - 3)/(2[x + 1][x - 1])
= (1 - x)/(2[x + 1][x - 1])
= (x - 1)/(- 2[x + 1][x - 1])
= - 1/(2[x + 1])
Answer: - 1/(2[x + 1])
= ([x + 2]/[{x + 1}{x - 1}]) - (3/[2{x - 1})
= (2x + 4 - 3x - 3)/(2[x + 1][x - 1])
= (1 - x)/(2[x + 1][x - 1])
= (x - 1)/(- 2[x + 1][x - 1])
= - 1/(2[x + 1])
Answer: - 1/(2[x + 1])
2009-08-18 10:08 pm
[(x + 2)/(x² - 1)] - [3/(2x - 2)]
= [(x + 2)/(x - 1)(x + 1)] - [3/2(x - 1)]
= [2(x + 2) - 3(x + 1]/2(x - 1)(x + 1)
= -(x - 1)/2(x - 1)(x + 1)
= -1/(2x + 1)
= [(x + 2)/(x - 1)(x + 1)] - [3/2(x - 1)]
= [2(x + 2) - 3(x + 1]/2(x - 1)(x + 1)
= -(x - 1)/2(x - 1)(x + 1)
= -1/(2x + 1)
2009-08-18 9:59 pm
Your answer is correct.
2009-08-21 1:49 am
( x + 2 ) / [ ( x - 1 ) ( x + 1 ) ] - (3/2) [ 1 / ( x - 1 ) ]
2x + 4 - 3 ( x + 1 )
------------------------------
2 ( x - 1 ) ( x + 1 )
1 - x
----------------------
2 ( x - 1 ) ( x + 1 )
- ( x - 1 )
-------------------------------
2 ( x - 1 ) ( x + 1 )
- 1
-------------------
2 ( x + 1 )
2x + 4 - 3 ( x + 1 )
------------------------------
2 ( x - 1 ) ( x + 1 )
1 - x
----------------------
2 ( x - 1 ) ( x + 1 )
- ( x - 1 )
-------------------------------
2 ( x - 1 ) ( x + 1 )
- 1
-------------------
2 ( x + 1 )
2009-08-19 12:40 am
(x + 2)/(x^2 - 1) - 3/(2x - 2)
= (x + 2)/[(x + 1)(x - 1)] - 3/[2(x - 1)]
= [2(x + 2)]/[2(x + 1)(x - 1)] - [3(x + 1)]/[2(x + 1)(x - 1)]
= (2x + 4)/[2(x + 1)(x - 1)] - (3x + 3)/[2(x + 1)(x - 1)]
= (2x + 4 - 3x - 3)/[2(x + 1)(x - 1)]
= (-x + 1)/[2(x + 1)(x - 1)]
= -(x - 1)/[2(x + 1)(x - 1)]
= -1/[2(x + 1)]
= (x + 2)/[(x + 1)(x - 1)] - 3/[2(x - 1)]
= [2(x + 2)]/[2(x + 1)(x - 1)] - [3(x + 1)]/[2(x + 1)(x - 1)]
= (2x + 4)/[2(x + 1)(x - 1)] - (3x + 3)/[2(x + 1)(x - 1)]
= (2x + 4 - 3x - 3)/[2(x + 1)(x - 1)]
= (-x + 1)/[2(x + 1)(x - 1)]
= -(x - 1)/[2(x + 1)(x - 1)]
= -1/[2(x + 1)]
2009-08-18 10:18 pm
There is no need for brackets around each fraction because division always comes before addition unless there are brackets.
(x + 2) / (x² - 1) - 3 / (2x - 2)
(x + 2) / (x + 1)(x - 1) - 3 / 2(x - 1)
2(x + 2) / 2(x + 1)(x - 1) - 3(x + 1) / 2(x + 1)(x - 1)
[2(x + 2) - 3(x + 1)] / 2[x + 1][x - 1]
[2x + 4 - 3x - 3] / 2[x + 1][x - 1]
(1 - x) / 2(x + 1)(x - 1)
-(x - 1) / 2(x + 1)(x - 1)
-1 / 2(x + 1)
-1 / (2x + 2)
(x + 2) / (x² - 1) - 3 / (2x - 2)
(x + 2) / (x + 1)(x - 1) - 3 / 2(x - 1)
2(x + 2) / 2(x + 1)(x - 1) - 3(x + 1) / 2(x + 1)(x - 1)
[2(x + 2) - 3(x + 1)] / 2[x + 1][x - 1]
[2x + 4 - 3x - 3] / 2[x + 1][x - 1]
(1 - x) / 2(x + 1)(x - 1)
-(x - 1) / 2(x + 1)(x - 1)
-1 / 2(x + 1)
-1 / (2x + 2)
2009-08-18 10:13 pm
[(x+2)/(x^2 -1)] - [3/(2x-2)] = [(x+2)/(x+1)(x-1)] - [3/2(x-1)]
= [(x+2)/(x+1)(x-1)] - [3(x+1)/2(x-1)(x+1)]
= [2(x+2)/2(x+1)(x-1)] - [3(x+1)/2(x-1)(x+1)]
= [(2x+4-3x-3)/2(x+1)(x-1)]
= -x+1/2(x+1)(x-1)
= 1-x/-2(x+1)(1-x)
= -1/2(x+1)
Check: Let x = 2.
[(2+2)/(2^2 -1)] - [3/(4-2)] = 4/3 - 3/2 = -1/6
For -1/2(x+1) = -1/2(2+1) = -1/6
= [(x+2)/(x+1)(x-1)] - [3(x+1)/2(x-1)(x+1)]
= [2(x+2)/2(x+1)(x-1)] - [3(x+1)/2(x-1)(x+1)]
= [(2x+4-3x-3)/2(x+1)(x-1)]
= -x+1/2(x+1)(x-1)
= 1-x/-2(x+1)(1-x)
= -1/2(x+1)
Check: Let x = 2.
[(2+2)/(2^2 -1)] - [3/(4-2)] = 4/3 - 3/2 = -1/6
For -1/2(x+1) = -1/2(2+1) = -1/6
2009-08-18 10:06 pm
x+2.........3
-------- - ---------
x^2-1.....2x-2
....x+2...............3
=-------------- - ------------- lcd 2(x+1)(x-1)
..(x+1)(x-1).....2(x-1)
...2(x+2)-3(x+1)
=-----------------------
.......2(x^2-1)
....2x+4-3x-3
=------------------
.......2(x^2-1)
....-x+1
=-------------
...2(x^2-1)
....-(x-1)
=----------------
..2(x+1)(x-1)
.....-1
=---------- answer//
..2(x+1)
-------- - ---------
x^2-1.....2x-2
....x+2...............3
=-------------- - ------------- lcd 2(x+1)(x-1)
..(x+1)(x-1).....2(x-1)
...2(x+2)-3(x+1)
=-----------------------
.......2(x^2-1)
....2x+4-3x-3
=------------------
.......2(x^2-1)
....-x+1
=-------------
...2(x^2-1)
....-(x-1)
=----------------
..2(x+1)(x-1)
.....-1
=---------- answer//
..2(x+1)
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