x²-y²+5x+3y+4
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x²-y²+5x+3y+4
(x+y-2xy)(x+y-z)+(xy-1)²
(x+y-2xy)(x+y-z)+(xy-1)²
(1)原式
=(x+y)(x-y)+(5x+3y)+4
=(x+y)(x-y)+(4x+4y)+x-y+4
=(x+y)(x-y+4)+(x-y+4)
=(x-y+4)(x+y+1)
第二题那个z应该是2吧
再问: 可不可以再帮我一个 加分
(x+y)的四次方-(x²-y²)²+(x-y)的四次方
再答:
(x+y)⁴-(x²-y²)²+(x-y)⁴
=(x+y)⁴+2(x²-y²)²+(x-y)⁴-3(x²-y²)²
=(x+y+x-y)²-3(x²-y²)²
=(2x)²-[√3(x²-y²)]²
=[2x+√3(x²-y²)][2x-√3(x²-y²)]
=(x+y)(x-y)+(5x+3y)+4
=(x+y)(x-y)+(4x+4y)+x-y+4
=(x+y)(x-y+4)+(x-y+4)
=(x-y+4)(x+y+1)
第二题那个z应该是2吧
再问: 可不可以再帮我一个 加分
(x+y)的四次方-(x²-y²)²+(x-y)的四次方
再答:
(x+y)⁴-(x²-y²)²+(x-y)⁴
=(x+y)⁴+2(x²-y²)²+(x-y)⁴-3(x²-y²)²
=(x+y+x-y)²-3(x²-y²)²
=(2x)²-[√3(x²-y²)]²
=[2x+√3(x²-y²)][2x-√3(x²-y²)]
(x+y)²+4(x+y)-5
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