10x平方 y平方-2x 4y 5＝0

81(x+y)²-121(m+n)²=[9(x+y)]²-[11(m+n)]²=(9x+9y)²-(11m+11n)²=[(9x+9y)+(

原式=(x+y)(x-y)/(x-y)^2=(x+y)/(x-y)=120/100=1.2^2指平方

(x平方-y平方)/(x+y)+(x平方-2xy+y平方)/(x-y)=(x+y)(x-y)/(x+y)+(x-y)^2/(x-y)=x-y+x-y=2(x-y)

2x+2y+z＝0(1)5x+4y-4z＝0(2)(2)-(1)*2得x=6z把x=6z代入（1）得y=-13z/2把x=6z,y=-13z/2代入3x平方-4y平方+5z平方／x平方+6y平方-10

15/16再问：要过程，这位大姐再答：由已知得X=﹣4／3Y，Z=﹣2／3Y，带入可得15／16再问：怎么得出再答：解方程组，①式*4+②式得x=-4/3y,代入①得z=-2/3y,再将x,z代入代数

X²-2X+Y²+6Y+10=0X²-2X+1+Y²+6Y+9=0（X-1）²+（Y+3）²=0X-1=0,X=1；Y+3=0,Y=-3（2

x^2+y^2-6x+2y+10=0(x^2-6x+9(+(y^2+2y+1)=0(x-3)^2+(y+1)^2=0因为(x-3)^2≥0,(y+1)^2≥0因此x-3=0,x=3y+1=0,y=-1

x平方+y的平方-4x-10y+29=0（x-2）²+（y-5）²=0要是它成立,必有x=2,y=5x平方y的平方+2x立方y平方+x的4次方y平=4×25+2×8×25+16×2

(3xy)平方/(-xy)+(x-2y)平方-(x+2y)(x-2y)=-9xy+(x-2y)(x-2y-x-2y)=-9xy+(x-2y)(-4y)=-9xy-4xy+8y²=8y

(3x平方-4y平方)+[-(x平方-2xy-y平方)]-[(3x平方-2xy-y平方)]=3x²-4y²-x²+2xy+y²-3x²+2xy+y&#

(y+1)平方-2x平方(1+y平方)+x四次方(1-y)平方=(y+1)²+2*x²*(y²-1)+[x²(y-1)]²-2x²(y

x平方+y平方+2xy分之x平方-4y平方除以x平方+xy分之2y+x=(x-2y)(x+2y)/(x+y)²·x(x+y)/(x+2y)=x(x-2y)/(x+y)

x²-4x+y²-10y+29=0x²-4x+4+y²-10y+25=0（x-2）²+（y-5）²=0∵（x-2）²≥0,（y-5

化简：(xy)^2-10xy+25+y^2-4y+4=0∴(xy-5)^2+(y-2)^2=0∴xy=5,y=2∴x=2.5代入,得：x^2y^2+2x^3*y^2+x^4*y^2=150+625/4

x平方+y平方-6x+2y+10=0,(x-3)^2+(y+1)^2=0因为：(x-3)^2>=0,(y+1)^2>=0所以：x-3=0,y+1=0x=3,y=-1x^2-y^2=8

【证法1】（梅文鼎证明）作四个全等的直角三角形,设它们的两条直角边长分别为a、b,斜边长为c.把它们拼成一个多边形,使D、E、F在一条直线上.过C作AC的延长线交DF于点P.∵D、E、F在一条直线上,

是2x吧x的平方-2x+y的平方-6y+10=0,x²-2x+1+y²-6y+9=0(x-1)²+(y-3)²=0x-1=0,y-3=0x=1,y=3再问：应该

=(x+y)(x+y)(x^2-xy+y^2)(x^2-xy+y^2)=(x^3+y^3)^2

[(3x-3y)/(y平方-x平方)+(x平方-2xy+y平方)/(x平方-y平方)-(x平方-x+6)/(x+2)]/(x+y)=【-3/(x+y)+(x-y)/(x+y)-(x-3)]】/(x+y

(1)(x²-y²)/(x²-2xy+y²)=(x-y)(x+y)/(x-y)²=(x+y)/(x-y)x²＋y²－2x－6y＋1