根号 12x²y 24x²y³

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求使 根号x+根号y

显然不等式两边都大于0平方后:x+y+2√xy≤a^2(x+y)整理下:(a^2-2)(x+y)+x+y-2√x≥0(a^2-2)(x+y)+(√x-√y)^2≥0若恒成立,显然需要满足(a^2-2)

根号x+2y×根号2x+4y

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

已知y=根号(x-8)+根号(8-x)+18,求代数式[(x+y)/(根号x+根号y)]-2xy/(x根号y-y根号x)

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

化简:((根号x-根号y)^3+2x根号x+y根号y)/(x根号x+y根号y)+(3根号xy-3y)/(x-y)

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)+[

解方程组x+y+根号下x+y=12 x+xy+y=23

设根号下x+y=a,则a平方+a=12可算出a1=-4,舍去a2=3所以x+y=9再和x+xy+y=23组成方程组,解得x1=7,y1=2x2=7y2=7

已知:丨2x+y-3丨+根号下(x-3y-5)²=0,则x²=?

解题思路:根据非负数的行政列出方程求出x,y的值即可求出x²的值解题过程:

X-Y除以根号X-根号Y等于?比较根号11减根号10和根号12减根号11的大小

(x-y)/(√x-√y)=(x-y)(√x+√y)/[(√x-√y)(√x+√y)]=(x-y)(√x+√y)/(x-y)=√x+√y)(√12-√11)²-(√11-√10)²

根号X*Y等于根号X*根号y吗?理由

当X,Y大于等于零时,根号X*Y等于根号X*根号y当X,Y小于零时,根号X*Y成立,根号X*根号y不存在

x根号x+x根号y/xy-y^2)-(x+根号xy+y/x根号x-y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]

若x+y=12,求根号(x^2+4)+根号(y^2+9)的最小值

高中的方法早忘了,给你介绍高数方法吧条件方程:x+y-12=0求最值方程:√(x^2+4)+√(y^2+9)则拉格朗日方程为L(x,y)=√(x^2+4)+√(y^2+9)+k(x+y-12)【k为某

已知x =2y 化简(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)-(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

若不等式根号x+根号y

√x+√y≤k√(x+y)平方得x+y+2√(xy)≤k²(x+y)∵2√(xy)≤x+y∴左≤2(x+y)恒成立,故有k²≥2,且显然k>0∴kmin=√2

代数式求值.已知x=2,y=根号3,求 (根号x-根号y)/(根号x+根号y)+(根号x+根号y)/(根号x-根号y)

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^2-2xy+y^2-根号3x-根号3y+12=0求x+y的最小值

x^2-2xy+y^2-√3x-√3y+12=0,令x=m+n,y=m-n代入化简得:4n^2-2√3m+12=02n^2-√3m+6=02n^2=√3m-6,m>=2√3xy=m^2-n^2=m^2

{(x-y)/(根号x+根号y)}-(x+y-2倍根号xy)/(根号x-根号y)=?

((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-√

已知x=2y,化简根号y/(根号x-根号y)-根号y/(根号x+根号y)

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)/

化简:(x-y)除以(根号x+根号y)-(x-2根号xy+y)除以(根号x-根号y)

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-√

根号x+y分之6除以根号x-y分之12

6/√(x+y)÷12/√(x-y)=√(x-y)/[2√(x+y)]=√(x^2-y^2)/[2(x+y)]