Question

先化简,后求值:(a-b/a+b)^2×(3a+3b/2a-2b)+a^2-b^2/ab,其中a=2,b=负1 【急~~~~~】

Answer 1

:(a-b/a+b)^2×(3a+3b/2a-2b)+a^2-b^2/ab,
=(a-b)^2/(a+b)^2×(3/2*(a+b)/(a-b)+a^2-b^2/ab
=3/2(a-b)/(a+b)+(a-b)(a+b)/ab
=1/2(a-b)(3/(a+b)+2(a+b)/ab)
=1/2(2+1)(3/(2-1)+2(2-1)/2*(-1))
=3/2(3-2)
=3/2

追问

你再检查下

追答

算错了，应是
(a-b/a+b)^2×(3a+3b/2a-2b)+a^2-b^2/ab,
=(a-b)^2/(a+b)^2×(3/2*(a+b)/(a-b)+a^2-b^2/ab
=3/2(a-b)/(a+b)+(a-b)(a+b)/ab
=1/2(a-b)(3/(a+b)+2(a+b)/ab)
=1/2(2+1)(3/(2-1)+2(2-1)/2*(-1))
=3/2(3-1)
=3

追问

恩恩、我算的也是3

Answer 2

自己做！！！！