设x2y2-20xy+x2+y2+81=0,求x+y的值
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/08 07:44:16
设x2y2-20xy+x2+y2+81=0,求x+y的值设x2y2-20xy+x2+y2+81=0,求x+y的值设x2y2-20xy+x2+y2+81=0,求x+y的值x2y2-20xy+x2+y2+
设x2y2-20xy+x2+y2+81=0,求x+y的值
设x2y2-20xy+x2+y2+81=0,求x+y的值
设x2y2-20xy+x2+y2+81=0,求x+y的值
x2y2-20xy+x2+y2+81=0x2y2-18xy+81+x2-2xy+y2=0(XY-9)的平方+(X-Y)=0XY-9=0 X-Y=0第一种 X=-3 Y=-3 所以x+y=-3-3=-6
第二种 X=3 Y=3 所以x+y=3+3=6
于是x+y=正负6
3.∵x2y2-20xy+x2+y2+81=0,
∴x2y2-18xy+81+x2-2xy+y2=0,
∴(xy-9)2+(x-y)2=0,
∴xy-9=0且x-y=0,
∴当x=3时,y=3;当x=-3时,y=-3.
x2y2-20xy+x2+y2+81=0 x^2y^2-18xy+81+x^2+y^2-2xy=0 (xy-9)^2+(x+y)^2=0 xy=9且x-y=0 x=3,y=3 所以x=y=6
x2y2-20xy+x2+y2+81=0 x2y2-18xy+81+x2-2xy+y2=0(xy-9)²+(x-y)²=0 xy-9=0 x-y=0 x=-3 y=-3或 x=3 y=3 所以x+y=6或x+y=-6
设x2y2-20xy+x2+y2+81=0,求x+y的值
x2+y2-x2y2-4xy-1
(x2+y2)(x2-xy+y2)-2x2y2因式分解
已知x2+4y2+x2y2+1=6xy,则(x4-y4)/(2x2+xy-y2)*(2x-y)/(xy-y2)/(x2+y2/y)2=
因式分解,x4-7x2y2+y4=?x4-7x2y2+y4=?应该是(x2+?xy+y2)(x2-?xy+y2) 的形式,那个?是多少
已知:x-y= -2 求(x2+y2)2-4xy(x2+y2)+4x2y2的值
x2y2-x2-y2+4xy+1分解因式
已知x2y2-20xy+x2+81=0求x,y的值
已知xy/x+y=1/3,x2y2/x2+y2=1/5,求xy的值.
分解因式(x2+y2)(x2-2xy+y2)+x2y2 注;2为平方
因式分解x2y2-x2-y2+1
X2y2-x2-y2+1因式分解.
已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/xy)2乘1/x+y的值
先化简,后求值,其中 x-y=1,xy=2,(1)x3y-2x2y2+xy3 (2)x2+y2
若实数x,y满足x2+y2+x2y2-4xy+1=0,则(x+y)2的值为
设x1,x2,y1,y2>0,a=根号(x1y1)+根号(x2y2),b=根号((x1+x2)(y1+y2)),则a,b的大小关
因式分解x2+y2-x2y2-4xy-1;
x2y2+xy-x2-y2+x+y+2分解因式,