標題:
MATHS LONG Q
發問:
最佳解答:
X2 - (Y-Z)2=0 X2 = (Y-Z)2 <===OK X = Y-Z or -X=Y-Z <===== square root results in positive or negative X - Y + Z =0 or X+Y-Z=0 You may use: [x-(Y-Z)][x+(Y-Z)]=0 X - Y + Z =0 or X+Y-Z=0 2X (N+1) - 7X (N) - 15X (N-1) [2X^2-7X-15]X^(N-1) [(2X+3)(X-5)]X^(N-1)
其他解答:
I think the correct answers are: X^2 - (Y-Z)^2 [Applying a^2-b^2 = (a+b)(a-b)] =[X+(Y-Z)][X-(Y-Z)] =(X+Y-Z)(X-Y+Z) 2X^(N+1) - 7X^N - 15X^(N-1) =X^(N-1) (2X^2 - 7X - 15) =X^(N-1) (2X+3)(X-5)
MATHS LONG Q
發問:
此文章來自奇摩知識+如有不便請留言告知
X2 - (Y-Z)2 X2 = (Y-Z)2 X = Y-Z X - Y + Z =0 (我可吾可以咁做?) 2X (N+1) - 7X (N) - 15X (N-1) (括號是次方)最佳解答:
X2 - (Y-Z)2=0 X2 = (Y-Z)2 <===OK X = Y-Z or -X=Y-Z <===== square root results in positive or negative X - Y + Z =0 or X+Y-Z=0 You may use: [x-(Y-Z)][x+(Y-Z)]=0 X - Y + Z =0 or X+Y-Z=0 2X (N+1) - 7X (N) - 15X (N-1) [2X^2-7X-15]X^(N-1) [(2X+3)(X-5)]X^(N-1)
其他解答:
I think the correct answers are: X^2 - (Y-Z)^2 [Applying a^2-b^2 = (a+b)(a-b)] =[X+(Y-Z)][X-(Y-Z)] =(X+Y-Z)(X-Y+Z) 2X^(N+1) - 7X^N - 15X^(N-1) =X^(N-1) (2X^2 - 7X - 15) =X^(N-1) (2X+3)(X-5)
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