题目内容
若实数a,b,c满足2a+2b=2a+b,2a+2b+2c=2a+b+c,则c的最大值是______.
由基本不等式得2a+2b≥2
=2×2
,即2a+b≥2
=2×2
,所以2a+b≥4,
令t=2a+b,由2a+2b+2c=2a+b+c可得2a+b+2c=2a+b2c,所以2c=
=1+
因为t≥4,所以1<
≤
,即1<2c≤
,所以0<c≤log2
=2-log23
故答案为:2-log23
| 2a2b |
| a+b |
| 2 |
| 2a2b |
| a+b |
| 2 |
令t=2a+b,由2a+2b+2c=2a+b+c可得2a+b+2c=2a+b2c,所以2c=
| t |
| t-1 |
| 1 |
| t-1 |
因为t≥4,所以1<
| t |
| t-1 |
| 4 |
| 3 |
| 4 |
| 3 |
| 4 |
| 3 |
故答案为:2-log23
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