arccosz 根号x^2 y^2的定义域

解∵x+2y≥0∴√(x+2y)×√(2x+4y)=√2√(x+2y)²=√2(x+2y)

y=根号(x-8)+根号(8-x)+18,x-8≥0,8-x≥0x=8,y=18[(x+y)/(根号x+根号y)]-2xy/(x根号y-y根号x)=26/(2√2+3√2)-288/(8*3√2-18

题是这样的吧：[(√x-√y)^3+2x√x+y√y]/(x√x+y√y)+[3√(xy)-3y]/(x-y)原式=[(x√x-3x√y+3y√x-y√y)+2x√x+y√y]/(x√x+y√y)+[

[2xy/（x·根号y+y·根号x）]=2根号xy的平方/[根号xy(根号x+根号y)]=2根号xy/(根号x+根号y)[x+y/(根号x+根号y)]+2根号xy/(根号x+根号y)=[(x+y+2根

根号x+根号y

根号内必须大于等于0故有x-1≥0且1-x≥0即x≥1且x≤1所以x=1将x=1代回去得y=3然后将x,y代入所求式即可你的所求式表述不是很清楚,所以没办法帮你求了

因为4x^2+9y^2-4x-6y+2=0,所以4x^2-4x+19y^2-6y+1=0,(2x-1)^2+(3y-1)^2=0所以2x-1=0,3y-1=0,所以x=1/2.y=1/3所以根号y/（

(x√x+x√y)/(xy-y^2)-[x+√(xy)+y]/(x√x-y√y)=[x(√x+√y)/[y(√x-√y)(√x+√y)]-[x+√(xy)+y]/{(√x-√y)[x+√(xy)+y]

结果为根号下x+根号下y解2xy/(x根号下y+y根号下x）分母提公因式根号下xy然后前后两式分母都含根号下x+根号下y合并后约分得根号下x+根号下y

额,题目很长,我读出二层意思,题中有2个/,哪一个最长啊?也就是哪个是哪个的被除数.

（根号y/根号x-根号y)-(根号y/根号x+根号y)=｛根号y（根号x+根号y）｝/（x-y）-｛根号y（根号x-根号y）｝/(x-y)=(y+y)/(x-y)因为x=2y所以原式=2y/y=2

原式=[(√x-√y)²+(√x+√y)²]/(√x+√y)(√x-√y)=(x+y-2√xy+x+y+2√xy)/(x-y)=2(x+y)/(x-y)=2(2+√3)/(2-√3

求定义域u=(arccosz)/√(x²+y²)定义域：-1≦z≦1,且x,y不同时为0；即由z=-1和z=1两张平行于xoy坐标面的平面所限定的空间【四周可无限扩张】,且不含z轴

（（x-y)/(√x+√y))-(x+y-2√xy)/(√x-√y),分母有理化,第一个式子分母乘以√x-√y,又(x+y-2√xy)=(√x-√y)(√x-√y),所以原式等于√x-√y-（√x-√

你的x、y分大小写吗?2Xy/x根号y什么意思?再问：没大小就是xyx十Y/（根号x十根号y）十2Xy/（x根号y十y根号x）再答：Y/（根号x十根号y）这一项分子分母都乘以（根号x-根号y）；2Xy

原式=√y/(√2y-√y)-√y/(√2y+√y)=√y/[√y(√2-1)]-√y/[√y(√2+1)]=1/(√2-1)-1/(√2+1)=(√2+1)/(√2+1)(√2-1)-(√2-1)/

1原式=X+Y/根号X+根号Y+(2根号XY/根号X+根号Y)=根号X加Y的二次方/根号X加根号Y=根号X+根号Y2原式=（a根号a+b根号b）/(a+b-根号ab）+（根号a+2）的2次方/（根号a

可知x≥0,y≥0(x-y)/(√x+√y)-(x-2√xy+y)/(√x-√y)=(√x+√y)(√x-√y)/(√x+√y)-(√x-√y)²/(√x-√y)=(√x-√y)-(√x-√