已知a+b=3 ab=2,求下列各式的值:1、a05b+ab05 2、a05+b05

1、a^0.5b+ab^0.5
=(ab)^0.5*(b^0.5+a^0.5)
=(ab)^0.5*[b+2(ab)^0.5+a]^0.5
=2^(0.5)*[3+2*2^(0.5)]^0.5
=(6+4根号2)^0.5
=2+根号2

2、a^0.5+b^0.5
=[a+2(ab)^0.5+b]^0.5
=[3+2*2^0.5]^0.5
=1+根号2

用二次项定展开,或自已一点展开:
(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5
=a^5+b^4+5ab(a^3+b^3)+10a^2b^2(a+b)
=a^5+b^4+5ab(a+b)(a^2+b^2-ab)+10a^2b^2(a+b)
=a^5+b^4+5ab(a+b)[(a+b)^2-3ab]+10a^2b^2(a+b)
代入a+b=3 ab=2
3^5=a^5+b^5+5*2*(3)(3^2-3*2)+10*2^2*(3)
a^5+b^5=3^5-5*2*(3)(3^2-3*2)-10*2^2*(3)
a^5+b^5=33

a05b+ab05=ab(a^4+b^4)=
(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4=a^4+b^4+4ab[(a+b)^2-2ab]+6a^2b^2
a^4+b^4=3^4-4*2(3^2-2*2)-6*2^2

a05b+ab05=ab(a^4+b^4)=2*(3^4-4*2(3^2-2*2)-6*2^2)=34

1.a²b+ab²=ab(a+b)=3×2=6
2. a²+b²=（a+b）²-2ab=9-4=5

题目错了吧，中间的那个0是什么？是0不会这样写的