先化简,在求值1.x²-(x²+2xy-y²)+5(y²-3xy),其中x=1/

1个回答

  • 1.原式=.x²-(x²+2xy-y²)+5(y²-3xy)

    =.x²-x²-2xy+y²+5y²-15xy

    =-2xy-15xy+y²+5y²

    =-17xy+6y²

    =Y(-17X+6Y)

    代入x=1/17,y=1/3

    原式=1/3*(-17*1/17+6*1/3)

    =1/3*(-1+2)

    =1/3*1

    =1/3

    2.原式=5xy-[x²+2xy-2y²﹚],

    =5xy-x²-2xy+2y²

    =5xy-2xy+2y²-x²

    =2y²-x²+3xy

    代入:x=﹣2,y=﹣1

    原式 =2*(-1)^2-(-2)^2+3*(-2)*(-1)

    =2-4+6

    =4

    补充:5xy-[x²+4xy-y²-(x²+2xy-2y²﹚],其中x=﹣2,y=﹣1

    原式=5xy-[x²+4xy-y²-x²-2xy+2y²]

    =5xy-[x²-x²+4xy-2xy+2y²-y²]

    =5xy-[2xy-y²]

    =5xy-2xy+y²

    =3xy+y²

    代入x=﹣2,y=﹣1 原式=3*(-2)*(-1)+(-1)^2

    =3*2+1

    =6+1

    =7