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有D Fatorization 數學吳識計, 幫幫拖!!!

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發問:

(x+y)^z^-4(x+y)(x-y)z^+2(x-y)^z^ (^=2次方) 點樣Fatorize? 幫幫拖! 請列出步驟!!!Thank!^_^

最佳解答:

(x+y)2z2-4(x+y)(x-y)z2+2(x-y)2z2 =z2[(x+y)2-4(x+y)(x-y)+2(x-y)2]..........抽公因數 =z2[(x+y)2-2(2)(x+y)(x-y)+2(x-y)2].........係吳係打錯野.? 2(x-y)2=4(x-y)2先得 =z2[(x+y)2-2(2)(x+y)(x-y)+4(x-y)2] =z2{(x+y)2-2(2)(x+y)(x-y)+[2(x-y)]2}.............a2b2=(ab)2 =z2{(x+y)-[2(x-y)]}2 =z2[x+y-(2x-2y)]2 =z2(x+y-2x+2y)2 =z2(3y-x)2 =[z(3y-x)]2.............a2b2=(ab)2

其他解答:

(x+y)2z2-4(x+y)(x-y)z2+2(x-y)2z2 =z2[(x+y)2-4(x+y)(x-y)+2(x-y)2]|||||你個fatorization串錯字,係factorization,同factorize.