x²-2x-2根号2x 3 2根号2-搜答案-搜搜考试网

首先须满足2x-1>=0,即x>=1/2方程右边mx>=0,所以也有m>=0方程两边平方,因[x+√(2x-1)][x-√(2x-1)]=x^2-2x+1=(x-1)^2,得：2x+2|x-1|=m^

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)+[

原式=2√3+√6-√36+30*√6/6=2√3+√6-6＋5√6=2√3+6√6-6

根号1-x^2,|x|

y=根号（1-x^2）所以y^2=1-x^2即x^2+y^2=1又因为y=根号（1-x^2）大于等于0-1

(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]

x+根号x+根号(x+2)+根号(x^2+2x)=根号x(根号x+1)+根号(x+2)(根号x+1)=(根号x+1)(根号x+根号(x+2))=3两边同时乘以（根号(x+2)-根号x）得(根号x+1)

1、x-2≥02-x≥0∴x=2不关于原点对称非奇非偶2、1-x^2≥0x^2-1≥0∴x=1或-1f(x)-f(-x)=0且f(x)+f(-x)=0所以既奇又偶3、x≠0f(x)+f(-x)=0∴奇

（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

√(x-2)/(x-2)÷√x/√(x³-2x²)=√(x-2)/(x-2)*√(x³-2x²)/√x=√(x-2)/(x-2)*x√(x-2)/√x=x(x-

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

答案为1.直接展开啊

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)/

（根号2-根号3)x

（-根20）×根2×根5=-根（20×2×5）=-根200=-10根2..

[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-

(√5+√3-√2)*(√5-√3+√2)=[√5+(√3-√2)]*[√5-(√3-√2)]=(√5)^2-(√3-√2)^2=5-(3+2-2√6)=2√6

因为根号(-x^2)有意义,则x=0所以答案为1-4+0+2=-1