Question

3)ChoE=-1/2xy^3+4xy-9 và F=5xy+4-1/2xy^3TinhE+F;E-F;F-E

Answer 1

\(E=-\dfrac{1}{2}xy^3+4xy-9\)

\(F=5xy+4-\dfrac{1}{2}xy^3=-\dfrac{1}{2}xy^3+5xy+4\)

\(E+F=-\dfrac{1}{2}xy^3+4xy-9+\left(-\dfrac{1}{2}xy^3\right)+5xy+4\)

        \(=-xy^3+9xy-5\)

\(E-F=-\dfrac{1}{2}xy^3+4xy-9-\left[\left(-\dfrac{1}{2}xy^3\right)+5xy+4\right]\)

        \(=-\dfrac{1}{2}xy^3+4xy-9+\dfrac{1}{2}xy^3-5xy-4\)

        \(=-xy-13\)

\(F-E=-\left(E-F\right)=-\left(-xy-13\right)\)

        \(=xy+13\)