Question

已知：x+3／x+2=1／根号3+根号2+1，求x-3／2x-4÷(5／x-2－x-2)的值

(x+3)/(x+2) = 1/(√3+√2+1)

Answer 1

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

原式=[(x-3)/(2x-4)]÷[5-(x+2)(x-2)]/(x-2)]
=[(x-3)/2(x-2)]×[-(x-2)/(x²-9)]
=[(x-3)/2(x-2)]×[-(x-2)/(x+3)(x-3)]
=-1/(x+3)
=√3+√2

Answer 2

楼上错误，答案应该是1/2(根号2+根号3)

Answer 3

因
(x+3)/(x+2)
=
1/(√3+√2+1)
所以
X+2
=
(X+3)(√3+√2+1)
＝（根号3＋根号2＋1）X＋3（根号3+根号2+1）
所以
X+3
=
-(√3-√2)
所以
，原式=
(X-3)/(2X-4)
*
(X-2)/(9-X^2)
=
-1/2(X+3)=(√3+√2)/2

Answer 4

把题写清楚一些，比如=后面的根号2+1,1是不是在根号下，这些都要说清楚