Question

已知 a＞b＞0，a²＋b²＝4ab，求a－b／a+b的值

妹妹的！求求大神了！

Answer 1

法一:
a－b／a+b
=(a+b-2b)/(a+b)
=1-2/(a/b+1)
∵a²＋b²＝4ab
∴(a/b)²+1=4a/b
∴(a/b)²-4a/b+1=0
△=4²-4=12
a/b=(4±2√3)/2=2±√3
∴原式=1-2/(2+√3+1)
=1-2/(3+√3)
=1-2(3-√3)/(9-3)
=1-(3-√3)/3
=√3/3
或原式=1-2/(2-√3+1)
=1-2/(3-√3)
=1-2(3+√3)/(9-3)
=1-(3+√3)/3
=-√3/3
∴原式=±√3/3
法二:(a-b)/(a+b)
=±√(a-b)²/(a+b)²
=±√(a²+b²-2ab)/(a²+b²+2ab)
=±√(2ab/6ab)
=±√3/3

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Answer 2

a>b>0 a-b>0 a+b>0
(a-b)^2/(a+b)^2
=(a^2+b^2-2ab)/(a^2+b^2+2ab)
=(4ab-2ab)/(6ab)
=1/3
所以(a-b)/(a+b)=根号3 /3

Answer 3

(a+b)^2=a^2+2ab+b^2=6ab
(a-b)^2=a^2-2ab+b^2=2ab
则（a-b)^2/(a-b)^2=1/3
a+b>0 a-b>0
(a-b)/(a+b)>0
(a-b)/(a+b)=根号3/3

Answer 4

（a-b）²=a²+b²-2ab=2ab
（a+b）²=a²+b²+2ab=6ab
a-b/a+b=√（（a-b/a+b）²）=√（a-b）²/（a+b）²=√（2ab/6ab）=√3/3