(10y+x)^2-(10x+y)=138 用一元2次方程解y=x+2

x,y 满足x(x-2)-y(y-6)=-10,求x,y的值 x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 (4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuaihuajian 3x+x+4y+(90+10)x=3y-(2x+y)+510y=34x {4x-3y-10=0 3x-2y=0,{3x-4(x-y)=2 2x-3y=1,{2(x+y)-(x-y)=3 (x+y)-2(x-y)=1 已知x²+4x+y²+6y+13=0,求x-y+(2y²/x+y) 请在10；20分前回答 若x-y不等于0,2x-3y=0,则分式(10x-11y)/(x-y)的值是（）?(10x-11y)/(x-y) =[(8x-8y)+(2x-3y)]/(x-y) =8为什么8X-8Y=8？(10x-11y)/(x-y)=[(8x-8y)+(2x-3y)]/(x-y)=8+(2x-3y)]/(x-y)=8+0=8 x+(9-8-y)=10 y+y-(9-8-y)=2 x-y=xx-9.9x xy=11y-1 y=? x=? x+y-xy=?x+(9-8-y)=10y+y-(9-8-y)=2x-y=xx-9.9xxy=11y-1 y=?x=?x+y-xy=? 已知2x-y=10,求式子[(x^2+y^2)-(x-y)^2+2y(x-y)]/4y (4xy-2y²)/4y是怎么变成=2y(2x-y)÷4y x^2+4y^2-6x+4y+10=0,求x^4-y^4/(x+2y)(x-2y)*x+2y/xy^2+y^3除以（x^2+y^2/y)^2的值 (x+y)^2/y^2-xy÷[-y^2+xy/(x-y)^2,x=0,y=根号10 x+2y=5 2x+y=10 已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y 就是已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y /4y就是除以4y x+5y=4,2x+10y=? 2x+4y=10 X*y=3 2x-y=0与x+y=10 25（x+y)^2+10(y-x)+1=?因式分解 25(x-y)^2+10(y-x)+1=