Question

已知X+3/X+2=1/根号3+根号2+1，求X-3/2X-4除以（5/X-2 -X-2）的值

Answer 1

取倒数
(x+2)/(x+3)=√3+√2+1
(x+2)/(x+3)-1=√3+√2+1-1
(x+2-x-3)/(x+3)=√3+√2
-1/(x+3)=√3+√2

原式
=[(x-3)/2(x-2)]/[(5/(x-2)-(x+2)]
=[(x-3)/2(x-2)]/[(5/(x-2)-(x+2)(x-2)/(x-2)]
=[(x-3)/2(x-2)]/[(5-x²+4)/(x-2)]
=[(x-3)/2(x-2)]/[-(x+3)(x-3)/(x-2)]
=-1/[2(x+3)]
=[-1/(x+3)]/2
=(√3+√2)/2

Answer 2

···········汗········你能加个扩号么·····我居然看不懂····1/根号3+根号2+1

Answer 3

确实问题应该再重新表述一下 哈哈~