2根号3x=根号2(x平方 1)

2x²-√3x-1=0x²-(√3/2)x=1/2x²-(√3/2)x+3/16=19/16(x-√3/4)²=19/16x=(√3+√19)/4或x=(√3-

[(2/3)x√(9x)+6x√(y/x)]+[y√(x/y)-x²√(1/x)]化简：原式=[(2/3)*3*x√x+6√(xy)]+[√(xy)-x√x]=2x√x+6√(xy)+√(x

根号2x平方-3X+根号2=0由十字相乘法进行因式分解得：(√2x-1)(x-√2)=0x=√2/2或x=√2

因为x=√3+1所以x﹥0,x-1﹥0原式=√[x/(1-2x+x)]=√[x/(x-1)]=x/(x-1)=(√3+1)/(√3+1-1)=(√3+1)/(√3)=[(√3+1)×√3]/(√3×√

1.先简化分母有理化x=2-√3y=2+√3x*x=7-4√3y*y=7+4√3所以x*x+y*y-2=12所以原式=2√32.x+y=√7xy=1/2（xxx+yyyy）（x+y）=（x+y）（xx

3√(x^2-2x+1)-√(x^2+4x+4)=3|x-1|-|x+2|={3(x-1)-(x+2)=2x-5,x>=1;{3(1-x)-(x+2)=1-4x,-2

(√3-√2)x²-(√3+√2)x=0x[(√3-√2)x-(√3+√2)]=0x=0,x=(√3+√2)/(√3-√2)所以x=0,x=5+2√6

即x²+(y-1)²=0则x=0,y-1=0x=0,y=1所以原式=(0+1+3)/(1*(0+1))=4

(√3+1)x^2-√3x+2-√3=0[(√3+1)x+1][x-(2-√3)]=0x=-1/(√3+1)或x=2-√3x=-(√3-1)/(√3+1)(√3-1)或x=2-√3x=-(√3-1)/

(1+根号2)X平方-(3+根号2)X+根号2=0利用十字相乘法得：[(1+根号2)x-1](x-根号2)=0x1=1/(1+根号2)=(根号2)-1x2=根号2

根号（(X+5)平方+Y平方）+根号（（X-5）平方+Y平方）=10,求X:根号[(x+5)²+y²]>=|x+5|,根号[(x-5)²+y²]>=|x-5|而

(x-3)²+√2(x-3)+1/2=3/2∴(x-3+√2/2)²=3/2∴x=3-√2/2+√6/2x=3-√2/2-√6/2

(根号3-根号2)x平方-1/根号3-根号2=0(√3-√2)x²=1/(√3-√2);x²=1/(√3-√2)²;x=±1/(√3-√2)=±(√3+√2)/(3-2)

√2x²-√3x+√3-√2=0(x-1)(√2x+√2-√3)=0x=1或x=√6/2-1

x²+(√6+√2)x+2√3=0所以△=b²-4ac=6+2√12+2-8√3=6-2√12+2=(√6-√2)²所以x=[-(√6+√2)±(√6-√2)]/2所以x

2x²-根号3x-1=0x=[根号3±√（3+4×2）]／（2×2）=(根号3±根号11)/4

由题意得：－1≤x≤2,则：√(x＋1)²＋√(x－3)²=|x＋1|＋|x－3|=(x＋1)＋(3－x)=4

等式两边同时平方,得12×3x=2(x平方+1)等式两边同时除以2,并将左边移到右边,得x平方-6x+1=0解得x=2根号2+3或-2根号2+3,两个解均满足x≥0(x在根号下),故均成立

√3(x²-x)-√2(x²+x)=0√3x(x-1)-√2x(x+1)=0x(√3x-√3-√2x-√2)=0x[(√3-√2)x-(√3+√2)]=0所以x1=0x2=(√3+

解√3(x²-1)=x(x-2)+1√3(x²-1)=x²-2x+1√3(x-1)(x+1)=(x-1)²当x=1时,显然成立当x≠1时方程两边除以x-1∴√3