已知2x²-4xy+5y²-12y+13＝1,则-搜答案-大师作文网

将其看出关于x的方程5x²-(6y+4)x+2y²+2y+1=0其判别式△≥0而△=(6y+4)²-20(2y²+2y+1)=-4y²+8y-4=-4

（3x-4y+2xy）-（2x-5y+5xy）=x+y-5xy=10

楼主是不是把题写错了呀,应该是（X+2)^2+/Y+1/=0吧因为（X+2)^2大于等于0/Y+1/大于等于0所以只有同时为0时,和才为0即X=-2Y=-1带入原式=4

（x+2）平方+|y+1|=0则x+2=0,y+1=0所以x=-2.y=-1xy平方=-2.x平方y=-45xy*-｛2x*y-【3xy*-（4xy*-2x*y）】｝=5*（-2）-【2*（-4）-3

根据已知得到x=-2y=-1原式=5xy^2-{2x^2y-[3xy^2-(4xy^2-2x^2y)]}=5xy^2-{2x^2y-[3xy^2-4xy^2+2x^2y]}=5xy^2-{2x^2y-

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-6xy+3x-3y=-6×(-2)+3×1=15

答：x+y=-1,xy=-2-5（x+y)+(x-y)+x(xy+y)=-5x-5y+x-y+xy(x+1)=-4x-6y+(-2)(x+1)=-4x-6y-2x-2=-6x-6y-2=-6(x+y)

x^2y+xy^2=xy(x+y)=1/5

原式=[4(x+y)-2xy]分之[(x+y)+xy]=[4(3xy)-2xy]分之[(3xy)+xy]=10xy分之2xy=5分之1

(3x－4y+2xy)－(2x－5y+5xy)=x+y-3xy=x+y-2xy-xy=(x-y)²-xy代入x-y=1,xy=－2得=1-(-2)=1+2=3

1.已知x+y=-4,xy=3,求5(xy/2+2y)+[6x-(2xy+y-3x)]的值5(xy/2+2y)+[6x-(2xy+y-3x)]=5xy/2+10y+6x-2xy-y+3x=xy/2+9

解2(x+3y-2xy)-(4x+8y-2xy)=2x+6y-4xy-4x-8y+8xy=-2x-2y+4xy=-2(x+y)+4xy=-2×5+4×3=12-10=24x²-3y²

-5（x+y)+(x-y)+2(xy+y)=-5x-5y+x-y+2xy+2y=-4x-4y+2xy=-4（x+y）+2xy=-4×（-1）+2×（-2）=4+（-4）=0你有问题也可以在这里向我提问

原式=-2x^2-2y^2-3xy-11xy=-2*7-14*(-1)=0

这道题目还是在考察韦达定理的运用用伟大定理求出xy的值再代入代数式否则是求不出来的(x+y)^2=x^2+y^2+2xy=1x^2+y^2=5(x-y)^2=5-2(-2)=9下面分两种情况讨论1x-

|x+2|+(y-1/2)的平方=0所以x+2=0y-1/2=0所以x=-2,y=1/2所以原式=4xy-x²-5xy+y²+x²+3xy=2xy+y²=-2+

因为xy/（x+y）=1/2所以x+y=2xy原式=3（x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1

5x^2-(3xy+4y^2)-(11xy-2y^2)-7x^2=5x^2-3xy-4y^2-11xy+2y^2-7x^2=-2x^2-2y^2-14xy=-2(x^2+y^2)-14*xy=-2*7

问题里的式子短符号的了

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6*(-2)+3*1=15不懂可追问,有帮助请采