Question

化简a根号a+b/a-b -b根号a-b/a+b -2b^2/根号a^2-b^2(a>b>0)

Answer 1

解：
原式=a√[(a+b)/(a-b)]-b√[(a-b)/(a+b)]-2b²/√(a²-b²) 分母有理化
=a√[(a+b)(a-b)/(a-b)²]-b√[(a+b)(a-b)/(a+b)²]-2b²√(a²-b²)/√(a²-b²)²
=[a√(a²-b²)]/(a-b)-[b√(a²-b²)]/(a+b)-[2b²√(a²-b²)]/(a²-b²) 通分
=[a(a+b)√(a²-b²)]/(a²-b²)-[b(a-b)√(a²-b²)]/(a²-b²)-[2b²√(a²-b²)]/(a²-b²)
={[a(a+b)-b(a-b)-2b²]√(a²-b²)}/(a²-b²)
=[(a²+ab-ab+b²-2b²)√(a²-b²)]/(a²-b²)
=[(a²-b²)√(a²-b²)]/(a²-b²)
=√(a²-b²)

Answer 2

已知a>b>0，那么：a+b>0，a-b>0
所以：
a根号[(a+b)/(a-b)] - b根号[(a-b)/(a+b)] - 2b^2/根号(a^2-b^2)
=a[根号(a²-b²)]/(a-b) -b[根号(a²-b²)]/(a+b) - 2b²[根号(a²-b²)]/(a²-b²)
=根号(a²-b²)*[ a/(a-b) -b/(a+b) - 2b²/(a²-b²)]
=根号(a²-b²)*[ (a²+ab-ab+b²- 2b²)/(a²-b²)]
=根号(a²-b²)*[ (a²-b²)/(a²-b²)]
=根号(a²-b²)

Answer 3