(1)[(2y+1)(2y-1)]^2 (2)(x+y)(x^2+y^2)(x-y)(x^4+y^4) (3)(2x+y+z)(y-2x-z)
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(1)[(2y+1)(2y-1)]^2 (2)(x+y)(x^2+y^2)(x-y)(x^4+y^4) (3)(2x+y+z)(y-2x-z)
解
[(2y+1)(2y-1)]²
=(4y²-1)²
=16y^4-8y²+1
(x+y)(x²+y²)(x-y)
=(x²-y²)(x²+y²)
=x^4-y^4
(2x+y+z)(y-2x-z)
=[y+(2x+z)][y-(2x+z)]
=y²-(2x+z)²
=y²-4x²-4xz-z²
(1)[(2y+1)(2y-1)]^2
=(4y²-1)²
=16y的4次方-8y²+1
(2) (x+y)(x^2+y^2)(x-y)(x^4+y^4)
=(x-y)(x+y)(x^2+y^2)(x^4+y^4)
=(x²-y²)(x²+y²)(x的4次方+y的4次方)
全部展开
(1)[(2y+1)(2y-1)]^2
=(4y²-1)²
=16y的4次方-8y²+1
(2) (x+y)(x^2+y^2)(x-y)(x^4+y^4)
=(x-y)(x+y)(x^2+y^2)(x^4+y^4)
=(x²-y²)(x²+y²)(x的4次方+y的4次方)
=(x的4次方-y的4次方)(x的4次方+y的4次方)
=x的8次方-y的8次方
(3)(2x+y+z)(y-2x-z)
=[y+(2x+z)][y-(2x+z)]
=y²-(2x+z)²
=y²-4x²-4xz-z²
收起
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