求证根号x^2 xy y^2 根号y^2 yz Z^2

把不等式右边的式子化成3/4(x+y)+3/4(x+z)+3/4(y+z)左边还是根号（x^2+xy+y^2）+根号（x^2+xz+z^2）+根号（y^2+yz+z^2)接下来分别证明根号（x^2+x

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)+[

任意做一个三角形ABC,并在三角形内部找到一点O,使得∠AOB=∠BOC=∠COA=120度,不妨设OA=x,OB=y,OC=z,在三角形AOB中,有余弦定理可得根号下(x^2+y^2-xy）=AB,

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

(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]

题目是√x^3+X^2y+1/4xy+√(1/4x^3)-X^2y+xy^2如果是：√x^3+X^2y+1/4xy+√(1/4x^3)-X^2y+xy^2=(3/2)√x^3+xy/4+xy^2=(3

（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

一张铁皮,第一次又去它的的8分之3,第二次用去它的5分之1,还剩这张铁皮的几分之几?

（（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-√

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

楼上两位都不对.证：x+y=1x+1/2+y+1/2=2[√(x+1/2)]²+[√(y+1/2)]²=2由均值不等式,得2[√(x+1/2)]²+[√(y+1/2)]&

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-√

[x+2√(x-1)]=[√(x-1)+1]^2[x-2√(x-1)]=[√(x-1)-1]^2x-1>=0x>=1y=√[x+2√(x-1)]+√[x-2√(x-1)]=√(x-1)+1+｜√(x-