已知下式,求a的整数部分:a=11×66+12×67+13×68+14×69+15×7011×65+12×66+13×67+14×68+15×69×1

2024-11-14 10:52:18
推荐回答(1个)
回答1:

∵分子:
11×66=(13-2)×(68-2)=13×68-2×13-2×68+4
12×67=(13-1)×(68-1)=13×68-13-68+1
13×68=13×68
14×69=(13+1)×(68+1)=13×68+13+68+1
15×70=(13+2)×(68+2)=13×68+2×13+2×68+4
∴11×66+12×67+13×68+14×69+15×70=13×68×5+10,
又∵分母:
11×65=(13-2)×(67-2),
12×66=(13-1)×(67-1),
13×67=13×67,
14×68=(13+1)×(67+1),
15×69=(13+2)×(67+2),
∴11×65+12×66+13×67+14×68+15×69=13×67×5+10,
a=

11×66+12×67+13×68+14×69+15×70
11×65+12×66+13×67+14×68+15×69
×100=
13×68×5+10
13×67×5+10
×100
∴a的整数部分是101.