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

Answer 1

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

追问

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）这步直接到X+3 = -(√3-√2)
我实在看不懂，请帮忙详细解释一下谢谢！

Answer 2

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-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