急诊专科护士:已知2x-y=10 求代数式[(x∧2+y∧2)-(x-y)∧2+2y(x-y)]÷4y的值

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[(x∧2+y∧2)-(x-y)∧2+2y(x-y)]÷4y
=(2xy+2xy-2y^2)/(4y)
=(2x-y)/2
=10/2
=5

[(x∧2+y∧2)-(x-y)∧2+2y(x-y)]÷4y
=(x∧2 + y∧2 - x∧2 + 2xy - y∧2 + 2xy - 2y∧2) ÷ 4y
=(4xy - 2y∧2) ÷ 4y
=(2x - y) ÷ 2
=5

｛（x^2+y^2）-（x-y）^2+2y（x-y）｝除4y
=(x²+y²-x²+2xy-y²+2xy-2y²)÷4y
=(4xy-2y²)/4y
=2y(2x-y)÷4y
=(2x-y)÷2
=10÷2
=5

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