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解:被除式是一个差的完全平方式
原式=(x²-4y²)²÷[(x+2y)²(x-2y)],被除式又是一个平方差式子
=[(x+2y)(x-2y)]²÷[(x+2y)²(x-2y)]
=(x+2y)²(x-2y)²÷[(x+2y)²(x-2y)]
=(x-2y)²÷(x-2y)
=x-2y
原式=(x²-4y²)²÷[(x+2y)²(x-2y)],被除式又是一个平方差式子
=[(x+2y)(x-2y)]²÷[(x+2y)²(x-2y)]
=(x+2y)²(x-2y)²÷[(x+2y)²(x-2y)]
=(x-2y)²÷(x-2y)
=x-2y
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(x^4-8x2y2+16y^4)÷(2y+x)2÷(x-2y)
=(x^2-4y^2)^2÷(2y+x)2÷(x-2y)
=[(x-2y)(x+2y)]^2÷(2y+x)2÷(x-2y)
=(x-2y)^2(x+2y)^2÷(2y+x)2÷(x-2y)
=[(x-2y)^2(x+2y)^2÷(2y+x)2]÷(x-2y)
=[(x+2y)(x-2y)^2]/2÷(x-2y)
=[(x-2y)(x+2y)]/2
=(x^2-4y^2)/2
=1/2 x^2-2y^2
=(x^2-4y^2)^2÷(2y+x)2÷(x-2y)
=[(x-2y)(x+2y)]^2÷(2y+x)2÷(x-2y)
=(x-2y)^2(x+2y)^2÷(2y+x)2÷(x-2y)
=[(x-2y)^2(x+2y)^2÷(2y+x)2]÷(x-2y)
=[(x+2y)(x-2y)^2]/2÷(x-2y)
=[(x-2y)(x+2y)]/2
=(x^2-4y^2)/2
=1/2 x^2-2y^2
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