y=1 2根号-x²+x+2
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![y=1 2根号-x²+x+2](/uploads/image/f/920520-0-0.jpg?t=y%EF%BC%9D1+2%E6%A0%B9%E5%8F%B7-x%C2%B2%2Bx%2B2)
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
根号(x+y-8)+根号(8-x-y)=根号(3x-y-4)+根号(x-2y+7),根据二次根式有意义得:X+Y-8≥0,8-X-Y≥0,∴X+Y≥8,X+Y≤8,∴X+Y=8,左边为0,右边两个非负
(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
((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)/
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(√x-√y)=3√y(5√y-√x)(√x)^2-√xy=15(√y)^2-3√xy(√x)^2+2√xy-15(√y)^2=0因式分解得:(√x+5√y)(√x-3√y)=0所以:√x+5√y
[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-
原方程可化减为(根x+根y)(根x-5根y)=0根x-5根y=0将x=5根y带入得:x+根号xy-y=29y2x+根号xy+3y=58y所以原式=29y/58y=1/2