已知a^2+b^2=5,ab=4求代数式5ab^2(a-b)-3ab(b-a)^2+5a^2b(b-a)的值 化解分式((a^2-1)⼀(a^2-2a+1)-1⼀...

(a-b)^2=a^2+b^2-2ab=5-2*4=-3(在实数范围内无解)
5ab^2(a-b)-3ab(b-a)^2+5a^2b(b-a)
=ab(a-b)[5b-3(a-b)-5a]
=ab(a-b)(8b-8a)
=-ab(a-b)^2
=12
((a^2-1)/(a^2-2a+1)-1/(a-1))/((a^2+a)/(a-1))
=[(a-1)(a+1)/(a-1)^2-1/(a-1)]*(a-1)/[a(a+1)]
=a/(a-1)*(a-1)/[a(a+1)]
=1/(a+1)

a^2+b^2=5,ab=4
则(a+b)^2=a^2+b^2+2ab=13, (a-b)^2=a^2+b^2-2ab=-3
5ab^2(a-b)-3ab(b-a)^2+5a^2b(b-a)
=5ab^2(a-b)-3ab(a-b)^2-5a^2b(a-b)
=ab(a-b)(5b-3a+3b-5a)
=ab(a-b)(8b-8a)
=-8ab(a-b)^2
=-8*4*9=-288

((a^2-1)/(a^2-2a+1)-1/(a-1))/((a^2+a)/(a-1))
=[(a+1)(a-1)/(a-1)^2-1/(a-1)]/[a(a+1)/(a-1)]
=[(a+1)/(a-1)-1/(a-1)]/[a(a+1)/(a-1)]
=[a/(a-1)]/[a(a+1)/(a-1)]
=1/(a+1)