（2007•宿迁）当a=2时，求（4a2−1-2a2+a）÷1a−1的值．

Question

（2007•宿迁）当a=√(2)时，求（(4)/(a2−1)-(2)/(a2+a)）÷(1)/(a−1)的值．

Answer 1

原式=[(4)/((a−1)(a+1))−(2)/(a(a+1))]÷(1)/(a−1)
=(4a−2(a−1))/(a(a−1)(a+1))×(a−1)
=(2(a+1))/(a(a+1))
=(2)/(a)
∵a=√(2)，
∴原式=(2)/(√(2))=√(2)．