Simplify the following algebraic fractors
１ 2x/1-x + 2/x-1
２ x+y/x-y - x-y/x+y
３ (2/a-3/5b) ÷ 2a-5b/12ab
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2007-01-29 13:04:52 · 4 個解答 · 發問者 比我做最佳回答呀 7 in 科學 ➔ 數學
請列出計算步驟
2007-01-29 13:10:53 · update #1
gd
2007-02-03 15:21:55 · answer #1 · answered by ShikaP 7 · 0⤊ 0⤋
1. 2x/1-x + 2/x-1
= 2x / (1-x) - 2 / (1-x)
= (2x - 2) / (1-x)
= 2 (x - 1) / (1 - x)
= -2
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2. x+y/x-y - x-y/x+y
= [ ( x + y )^2 - ( x - y )^2 ] / ( x^2 - y^2 )
= [ ( x + y + x - y ) * ( x + y - x + y ) ] / ( x^2 - y^2 )
= 2x2y / ( x^2 - y^2 )
= 4xy / ( x^2 - y^2 )
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3. (2/a-3/5b) ÷ 2a-5b/12ab
= [ ( 10 b - 3 a ) / 5ab ] * ( 12 ab) / ( 2 a - 5 b )
= (120b - 36a) / (10a - 25b)
2007-01-31 17:01:26 · answer #2 · answered by ? 6 · 0⤊ 0⤋
2x/1-x + 2/x -1
=2/(x-1) - -〔2x/(x-1)〕
=(2--2x)/(x-1)
=3x/(x-1)
x+y/x-y - x-y/x+y
=｛〔(x+y)(x-y)〕/ 〔(x^2)-(y^2)〕｝- ｛〔(x+y)(x-y)〕/ 〔(x^2)-(y^2)〕｝
= ｛〔(x^2)-(y^2)〕/〔(x^2)-(y^2)〕｝-｛〔(x^2)-(y^2)〕/〔(x^2)-(y^2)〕｝
= 1-1
= 0
(2/a-3/5b) ÷ 2a-5b/12ab
=〔(2/a)-(3/5b) 〕 × 12ab/(2a-5b)
= 〔(10b-3a)/5ab〕× 12ab/(2a-5b)
= 〔120a(b^2)-36(a^2)b〕/ 〔10(a^2)b-25a(b^2)〕
2007-01-29 13:23:46 · answer #3 · answered by 平昔鳥 5 · 0⤊ 0⤋
1.=x*
2=y
3=a÷ b
2007-01-29 13:08:32 · answer #4 · answered by bckps6711 3 · 0⤊ 0⤋