(2x∧2-4x-10xy)/(?)=(1/2)x-1-(5/2)y

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(2x∧2-4x-10xy)/(?)=(1/2)x-1-(5/2)y(2x∧2-4x-10xy)/(?)=(1/2)x-1-(5/2)y(2x∧2-4x-10xy)/(?)=(1/2)x-1-(5/2

(2x∧2-4x-10xy)/(?)=(1/2)x-1-(5/2)y
(2x∧2-4x-10xy)/(?)=(1/2)x-1-(5/2)y

(2x∧2-4x-10xy)/(?)=(1/2)x-1-(5/2)y
(2x∧2-4x-10xy)/(4x)=(1/2)x-1-(5/2)y