分解因式(a²+b²)²-(2ab)² (x+y)²-2(x²-y²)+(x-y)²
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分解因式(a²+b²)²-(2ab)² (x+y)²-2(x²-y²)+(x-y)²
分解因式(a²+b²)²-(2ab)² (x+y)²-2(x²-y²)+(x-y)²
分解因式(a²+b²)²-(2ab)² (x+y)²-2(x²-y²)+(x-y)²
(a²+b²)²-(2ab)²
=[(a²+b²)+2ab][(a²+b²)-2ab]
=(a²+2ab+b²)(a²-2ab+b²)
=(a+b)²(a-b)²
(x+y)²-2(x²-y²)+(x-y)²
=(x+y)²-2(x+y)(x-y)+(x-y)²
=[(x+y)-(x-y)]²
=(2y)²
=4y²
答:
(a²+b²)²-(2ab)²
=(a²+b²-2ab)(a²+b²+2ab)
=(a-b)(a-b)(a+b)(a+b)
(x+y)²-2(x²-y²)+(x-y)²
=x²+2xy+y²-2x²+2y²+x²-2xy+y²
=4y²
(a²+b²)²-(2ab)²
=﹙a²+b²+2ab﹚﹙a²+b²-2ab﹚
=﹙a+b﹚²﹙a-b﹚²
(x+y)²-2(x²-y²)+(x-y)²
=(x+y)²-2(x+y)﹙x-y﹚+(x-y)²<...
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(a²+b²)²-(2ab)²
=﹙a²+b²+2ab﹚﹙a²+b²-2ab﹚
=﹙a+b﹚²﹙a-b﹚²
(x+y)²-2(x²-y²)+(x-y)²
=(x+y)²-2(x+y)﹙x-y﹚+(x-y)²
=[(x+y)-(x-y)]²
=﹙x+y-x+y﹚²
=(2y)²
=4y²
求采纳~~~~~~~~~
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