已知:x=2分之根号3减1,求x平方减x加1-搜答案-智慧作业帮

∵x=√3-1∴x/(x-1)=(√3-1)/(√3-2)=(√3-1)(√3+2)/(3-4)=-(3-2+√3)=-(√3+1)∴x²/(x²-2x+1)=x²/(x

因为x=√3+1所以x﹥0,x-1﹥0原式=√[x²/(1-2x+x²)]=√[x²/(x-1)²]=x/(x-1)=(√3+1)/(√3+1-1)=(√3+1

（x*x+1）/x=x+1/x因为根号x+根号1/x=3,所以等式两边同时平方得x+1/x+2=9,所以x+1/x=9-2=7

已知x=根号2+1分之1,x=1/(√2+1)x=√2-1代数式3-根号x的平方-4x+4=3-√(x-2)^2=3-(2-x)=1+x=1+√2-1=√2

x=根号3+1分之2=(3-1)分之2(根号3-1)=根号3-1x²+2x-1=x(x+2)-1=(根号3-1)(根号3+1)-1=3-1-1=1再问：把下列各式分母有理化：（1）（根号a-

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-1)/2]-(x/3)=13(x-1)-2x=6x=9√x/(x+2)=3/11

x=(根号3-根号2)/(根号3+根号2)=(根号3-根号2)^2/[(根号3)^2-(根号2)^2]=(根号3-根号2)^2=5-2*根号6类似可得y=5+2*根号6xy=1,x^2+y^2=(5-

x=2/(1-√3)=2(1+√3)/[(1+√3)(1-√3)]=2(1+√3)/(1²-√3²)=2(1+√3)/(1-3)=2(1+√3)/(-2)=-1-√3x²

x=根号3+根号2分之根号3-根号2=(√3-√2)²y=根号3-根号2分之根号3+根号2=(√3+√2)²∴xy=(√3-√2)²(√3+√2)²=1x-y=

2x/(x^2-1)-1/(x+1)=2x/(x^2-1)-(x-1)/(x^2-1)=(2x-x+1)/(x^2-1)=(x+1)/(x^2-1)=1/(x-1)=1/(根号3+1-1)=1/(根号

x=2/(√3-1)=2(√3+1)/(3-1)=√3+1所以x-1=√3x²-2x+2=(x-1)²+1=√3²+1=3+1=4

根号X-根号X分之一=2二边平方得:x-2根号X*根号1/X+1/x=4即X+1/X=4+2=6平方得:X^2+2+1/X^2=36所以,X^2+1/X^2=34

(X+2)/(X²-X-6)=1/(X-3)=1/[(√2-1)/3-3]=-(30+3√2)/8再问：是x+2分之x^2-x-6再答：(X+2)/(X²-X-6)=(X-3)(X

X^2-3X+1=0x+1/x=3x+2+1/x=5（根号x+根号x分之1）平方=5则根号x+根号x分之1=根号5再问：还有若根号(2012-m)²+根号m-2013=m,求代数式m-201

x=2分之根号3-根号2y=2分之根号3+根号2xy=1/4y-x=根号2x分之1-y分之1=(y-x)/xy=根号2/(1/4)=4根号2

依题意得x+x-3=根号22x-3=根号22x=根号2+3同时除以2得x=根号2/2+3/2

x=1/(2-√3)=2+√3,y=1/(2+√3)=2-√3则xy=1,x+y=(2+√3)+(2-√3)=410x的平方+xy+10y的平方=10(x^2+y^2)+1=10(x+y)^2-20x

x-1=1-3^(1/2)