Question

已知a+b=5，ab=3，求a的三次方b+2a的平方b的平方+ab的三次方

计算

Answer 1

a+b=5
ab=3
a+3/a=5
a^2+3=5a
变形为a^2-5a+3=0
公式a=[5±√(25-12)]÷2=(5±√13)/2=2.5±√13/2
b=2.5-±√13/2
由于两组解最后算出的答案一样，任取一组即可：取a1=2.5+√13/2
b1=2.5-√13/2
a^2=2.5^2+(√13/2)^2+2×2.5×√13/2=6.25+3.25+2.5√13=9.5+2.5√13
b^2=2.5^2+(√13/2)^2-2×2.5×√13/2=6.25+3.25-2.5√13=9.5-2.5√13

a^3b+2×a^2×b^2+ab^3
=3a^2+3b^2+2×a^2×b^2
=3×(9.5+2.5√13)+3×(9.5-2.5√13)+2×(9.5+2.5√13)(9.5-2.5√13)
=3×9.5+3×9.5+2×[(9.5)^2-(2.5√13)^2]
=57+2×(90.25-6.25×13)
=57+18
=75

Answer 2

原式=ab(a²+2ab+b²)
=ab(a+b)²
=3×5²
=75

Answer 3

所求= ab（a^2 + 2ab+ b^2）= ab（a+b）^2 = 3 * 25 = 75

Answer 4

=ab*(a+b)^2
=75