第24页

信息发布者:
$x$
$2b$
$12ab$
$\frac{1}{4}y^2$
$a$
$\frac{1}{2}b$
$\frac{1}{4}b^2$
解:原式= $y^2+6y+9$
解:原式=$9x^2-6x+1$
解:原式=$\frac{1}{4}x^2-\frac{1}{3}xy+\frac{1}{9}y^2$
解:原式=$4x^2-4x+1$
$\pm6$
D
D
解:
(1)由题意得:
$\begin{cases}(x-y)^2 = x^2 - 2xy + y^2 = 4 & ① \\(x+y)^2 = x^2 + 2xy + y^2 = 16 & ②\end{cases}$
①+②得:$2(x^2 + y^2) = 20,$
解得$x^2 + y^2 = 10。$
(2)②-①得:$4xy = 12,$
解得$xy = 3$