初二 数学 二次根式 请详细解答,谢谢! (20 15:6:51)

来源：百度知道 编辑：UC知道 时间：2024/06/06 21:12:10
已知  x+3/x+2=1/√3+√2+1,求  x-3/2x-4÷((5/x-2)-x-2)的值
   /为分号        √为根号

∵x+3/x+2=1/√3+√2+1 ,
∴X+2/x+3=√3+√2+1 ,
即[(x+3)-1)/(x+3)==√3+√2+1
∴-1/(x+3)=√3+√2
x-3/2x-4÷(5/x-2 -x-2)
=(x-3)/2(x-2)÷(9-x^2)/(x-2)
=(x-3)/2(x-2)*(x-2)/(3+x)(3-x)
=-1/2*1/(x+3)
=(√3+√2)/2

∵x+3/x+2=1/√3+√2+1 ,
∴X+2/x+3=√3+√2+1 ,
即[(x+3)-1)/(x+3)==√3+√2+1
∴-1/(x+3)=√3+√2
x-3/2x-4÷(5/x-2 -x-2)
=(x-3)/2(x-2)÷(9-x^2)/(x-2)
=-1/2*1/(x+3)
=(√3+√2)/2
就OK了

其实很简单的,方法见下：
x+3/x+2=1/√3+√2+1 ,
X+2/x+3=√3+√2+1 ,
[(x+3)-1)/(x+3)=√3+√2+1
-1/(x+3)=√3+√2

x-3/2x-4÷(5/x-2 -x-2)
=(x-3)/2(x-2)÷(9-x^2)/(x-2)
=-1/2*1/(x+3)
=(√3+√2)/2