1.x³-x²y-xy²+y³ 2.y²-2y-15 3.已知a+b=2,ab=2,求1/2a³b+a²b²+1/2ab³

1.x³-x²y-xy²+y³ 2.y²-2y-15 3.已知a+b=2,ab=2,求1/2a³b+a²b²+1/2ab³
1.x³-x²y-xy²+y³ 2.y²-2y-15 3.已知a+b=2,ab=2,求1/2a³b+a²b²+1/2ab³

1.x³-x²y-xy²+y³ 2.y²-2y-15 3.已知a+b=2,ab=2,求1/2a³b+a²b²+1/2ab³
1.x³-x²y-xy²+y³
=x(x²-y²)-y(x²-y²)
=(x²-y²)(x-y)
=(x-y)²(x+y)
2.y²-2y-15
=(y-5)(y+3)
3.已知a+b=2,ab=2,求
1/2a³b+a²b²+1/2ab³
=1/2ab(a²+2ab+b²)
=1/2ab(a+b)²
=1/2×2×2²
=4

1.x³-x²y-xy²+y³
=(x³-x²y)-(xy²-y³ )
=x²(x-y)-y²(x-y)
=(x-y)(x²-y²)
=(x-y)(x+y)(x-y)
=(x-y)²(x+y)
2.y²-2y-1...

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1.x³-x²y-xy²+y³
=(x³-x²y)-(xy²-y³ )
=x²(x-y)-y²(x-y)
=(x-y)(x²-y²)
=(x-y)(x+y)(x-y)
=(x-y)²(x+y)
2.y²-2y-15
=(y-5)(y+3)
3.已知a+b=2，ab=2，求
1/2a³b+a²b²+1/2ab³
=1/2ab(a²+2ab+b²)
=1/2ab(a+b)²
=1/2×2×2²
=4

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1.原=（x³-x²y）-（xy²-y³）
=x²（x-y）-y²（x-y）
=（x²-y²）（x-y）
2.原方程整理得（y+3）（y-5）=0
∴x1=-3，x2=5
3.原=1/2ab（a²+b²）+a²b²...

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1.原=（x³-x²y）-（xy²-y³）
=x²（x-y）-y²（x-y）
=（x²-y²）（x-y）
2.原方程整理得（y+3）（y-5）=0
∴x1=-3，x2=5
3.原=1/2ab（a²+b²）+a²b²
=（a²+b²）+4
=（a+b）²-2ab+4
=4-4+4
=4

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