| 1 |
| 4 |
| 1 |
| 4 |
即（x+2y-
| 1 |
| 2 |
∴x+2y-
| 1 |
| 2 |
則x+2y=
| 1 |
| 2 |
故答案為：
| 1 |
| 2 |
已知實(shí)數(shù)x，y滿足x2+4xy+4y2-x-2y+1/4=0，則x+2y的值為 _ ．
已知實(shí)數(shù)x，y滿足x2+4xy+4y2-x-2y+
=0，則x+2y的值為 ___ ．
| 1 |
| 4 |
數(shù)學(xué)人氣：782 ℃時(shí)間：2019-10-30 03:32:12
優(yōu)質(zhì)解答
已知等式變形得：x2+4xy+4y2-x-2y+
=（x+2y）2-（x+2y）+
=0，
即（x+2y-
）2=0，
∴x+2y-
=0，
則x+2y=
．
故答案為：
即（x+2y-
∴x+2y-
則x+2y=
故答案為：
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