標題:
數學題001.......因式分解
發問:
展開(x+1)(x+4)和(x+2)(x+3) 因式分解(x+1)(x+2)(x+3)(x+4)-48 更新: sorry it is 展開(x-1)(x-4)和(x-2)(x-3) 因式分解(x-1)(x-2)(x-3)(x-4)-48
最佳解答:
(x-1)(x-4)=x^2-5x+4 (x-2)(x-3)= x^2-5x+6 (x-1)(x-2)(x-3)(x-4)-48 =(x^2-5x+4)(x^2-5x+6)-48 =[(x^2-5x+5)-1][(x^2-5x+5)+1]-48 =(x^2-5x+5)^2-1^2-48 =(x^2-5x+5)^2-49 =(x^2-5x+5)^2-7^2 =[(x^2-5x+5)+7] [(x^2-5x+5)-7] =(x^2-5x+12)(x^2-5x-2)
其他解答:
數學題001.......因式分解
發問:
展開(x+1)(x+4)和(x+2)(x+3) 因式分解(x+1)(x+2)(x+3)(x+4)-48 更新: sorry it is 展開(x-1)(x-4)和(x-2)(x-3) 因式分解(x-1)(x-2)(x-3)(x-4)-48
最佳解答:
(x-1)(x-4)=x^2-5x+4 (x-2)(x-3)= x^2-5x+6 (x-1)(x-2)(x-3)(x-4)-48 =(x^2-5x+4)(x^2-5x+6)-48 =[(x^2-5x+5)-1][(x^2-5x+5)+1]-48 =(x^2-5x+5)^2-1^2-48 =(x^2-5x+5)^2-49 =(x^2-5x+5)^2-7^2 =[(x^2-5x+5)+7] [(x^2-5x+5)-7] =(x^2-5x+12)(x^2-5x-2)
其他解答:
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(x+1)(x+4)和(x+2)(x+3)=(x+1)(x+4)和(x+2)(x+3) (x+1)(x+2)(x+3)(x+4)-48 =(x+1)(x+2)(x+3)(x+4)-48文章標籤
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