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82问答网 > 已知X+3⼀X+2=1⼀根号3+根号2+1，求X-3⼀2X-4除以（5⼀X-2 -X-2）的值

已知X+3⼀X+2=1⼀根号3+根号2+1，求X-3⼀2X-4除以（5⼀X-2 -X-2）的值

2024-12-04 15:22:56

推荐回答（3个）

回答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

回答2：

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

回答3：

确实问题应该再重新表述一下哈哈~

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