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已知x+y=3,xy=1,求x^2+y^2;(x-y)^2;x^3y^2+x^2y^3的值
3个回答
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(x+y)²=3²
x²+2xy+y²=9
x²+y²=9-2xy
=9-2×1
=7
(x-y)²
=x²-2xy+y²
=7-2×1
=5
x³y²+x²y³
=x²y²(x+y)
=1²×3
=3
x²+2xy+y²=9
x²+y²=9-2xy
=9-2×1
=7
(x-y)²
=x²-2xy+y²
=7-2×1
=5
x³y²+x²y³
=x²y²(x+y)
=1²×3
=3
展开全部
x^2+y^2=(x+y)^2-2xy=7
(x-y)^2=(x+y)^2-4xy=5
x^3y^2+x^2y^3
=x^2y^2(x+y)
=3
(x-y)^2=(x+y)^2-4xy=5
x^3y^2+x^2y^3
=x^2y^2(x+y)
=3
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展开全部
x^2+y^2=(x+y)²-2xy=3²-2=7
(x-y)^2=(x+y)²-4xy=3²-4=5
x^3y^2+x^2y^3=(xy)²(x+y)=1²×3=3
(x-y)^2=(x+y)²-4xy=3²-4=5
x^3y^2+x^2y^3=(xy)²(x+y)=1²×3=3
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