標題:
MATHS(f.2) only one question
發問:
最佳解答:
49 ? (2x ?1)2 = [7 + (2x ?1)][7 ? (2x ?1)] = (2x + 6)(8 ? 2x ) ==> (2x + 6) = 2( x + 3) ==> (8 ? 2x ) = 2 (4 ? x ) = 2( x + 3) X 2 (4 ? x ) = (2X2)( x + 3)(4 ? x ) = 4( x + 3)(4 ? x )
其他解答:
首先,49=7^2 然後,將條式看作 (7)^2 * (2x-1)^2 Then,用 formula a^2 * b^2 = (a+b)*(a-b) 做代入法 變做 [ 7+ (2x - 1) ] * [ 7- (2x - 1) ] 之後計...... (7+2x-1)(7-2x+1) ^依道要負負得正 =(2x+6)(8-x) 再抽 common factor (2) =4[ (x+3) * (4-x) ] =4 (x+3) (4-x) 明未? 2007-10-29 15:32:29 補充： 負負得正係[7-(2x-1)]果道|||||[7-(2x-1)][7+(2x-1)] =(8-2x)(6+2x) =2(4-x)2(3+x) =4(4-x)(3+x)|||||49 ? (2x ?1)^2 =7^2-(2x-1)^2 =7^2-(2x)^2-2(2x 1)+1^2 =7^2-(2x)^2-4x+1 =(7+2x)(7-2x)-4x+1
MATHS(f.2) only one question
發問:
此文章來自奇摩知識+如有不便請留言告知
49 ? (2x ?1)2 = [7 + (2x ?1)][7 ? (2x ?1)] = (2x + 6)(8 ? 2x ) = 4( x + 3)(4 ? x ) why 4( x + 3)(4 ? x )?點得出????最佳解答:
49 ? (2x ?1)2 = [7 + (2x ?1)][7 ? (2x ?1)] = (2x + 6)(8 ? 2x ) ==> (2x + 6) = 2( x + 3) ==> (8 ? 2x ) = 2 (4 ? x ) = 2( x + 3) X 2 (4 ? x ) = (2X2)( x + 3)(4 ? x ) = 4( x + 3)(4 ? x )
其他解答:
首先,49=7^2 然後,將條式看作 (7)^2 * (2x-1)^2 Then,用 formula a^2 * b^2 = (a+b)*(a-b) 做代入法 變做 [ 7+ (2x - 1) ] * [ 7- (2x - 1) ] 之後計...... (7+2x-1)(7-2x+1) ^依道要負負得正 =(2x+6)(8-x) 再抽 common factor (2) =4[ (x+3) * (4-x) ] =4 (x+3) (4-x) 明未? 2007-10-29 15:32:29 補充： 負負得正係[7-(2x-1)]果道|||||[7-(2x-1)][7+(2x-1)] =(8-2x)(6+2x) =2(4-x)2(3+x) =4(4-x)(3+x)|||||49 ? (2x ?1)^2 =7^2-(2x-1)^2 =7^2-(2x)^2-2(2x 1)+1^2 =7^2-(2x)^2-4x+1 =(7+2x)(7-2x)-4x+1
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