Question

初一数学题..............速度

键盘的括号键坏了，用{}代表
已知|x+2|+{y-1/2}^2=0，求代数式4xy-{x^2+5xy-y^2}+{x^2+3xy}的值。

Answer 1

【参考答案】

根据题意，l x+2 l=(y- 1/2)²=0
解得 x=-2, y=1/2

4xy-{x^2+5xy-y^2}+{x^2+3xy}
=4xy-x²-5xy+y²+x²+3xy
=2xy+y²
=2×(-2)×(1/2)+(1/4)
=-2+(1/4)
=-7/4

Answer 2

由|x+2|+{y-1/2}^2=0 得 x=-2,y=1/2
4xy-{x^2+5xy-y^2}+{x^2+3xy}
=4xy-x^2-5xy+y^2+x^2+3xy
=2xy+y^2
代入x,y 原式=2xy+y^2＝-2+1/4=-7/4

Answer 3

x=-2 y=1/2

4xy-(x^2+5xy-y^2)+(x^2+3xy)

=2xy+y²
=-2+1/4
=-7/4

Answer 4

因为|x+2|、(y-1/2)²不可能为负数，且|x+2|+( y -1/2)²=0,
所以|x+2|=0、(y -1/2)²=0
x+2=0、y -1/2=0
x=-2，y=1/2

所以4XY-(X^2+5xy-y^2)+(x^2+3xy)
=-4+1+1/4+1
=-2+1/4
=-1.75

希望楼主满意。

Answer 5

由条件可知，x=-2，y=1/2，那么
4xy-{x^2+5xy-y^2}+{x^2+3xy}=4xy-x^2-5xy+y^2+x^2+3xy=y^2+2xy=1/4+2*(-2)*1/2=7/4

Answer 6

由前提得：x+2=0 x=-2 y-1/2=0 y=1/2
4xy-（x^2+5xy-y^2）+(x^2+3xy)=2xy-y^2=2*(1/2)*(-2)-(1/2)^2=-2-(1/4)=-9/4