
解;令ab-b+a=k,则ab=k-(a-b).
(a-b)²=a²+b²-2ab=9-2[k-(a-b)]
=9-2k+2(a-b).
∴(a-b)²-2(a-b)+2k-9=0.
=(-2)²-4(2k-9)=40-8k≧0.
8k≦40, k≦5. 即:ab-b+a的最大值为5.
解;令ab-b+a=k,则ab=k-(a-b).
(a-b)²=a²+b²-2ab=9-2[k-(a-b)]
=9-2k+2(a-b).
∴(a-b)²-2(a-b)+2k-9=0.
=(-2)²-4(2k-9)=40-8k≧0.
8k≦40, k≦5. 即:ab-b+a的最大值为5.