典例1
解 (1)$(2x+3y)^2$
$=(2x)^2 + 2·(2x)·(3y)+(3y)^2$
$=4x^2 +12xy +9y^2.$
(2)$(\frac{1}{3}m - \frac{1}{2}n)^2$
$=(\frac{1}{3}m)^2 -2·(\frac{1}{3}m)·(\frac{1}{2}n)+(\frac{1}{2}n)^2$
$=\frac{1}{9}m^2 - \frac{1}{3}mn + \frac{1}{4}n^2.$
(3)$(3x - y)^2 - (x - 2y)(x + 2y)$
$=9x^2 -6xy + y^2 -x^2 +4y^2$
$=8x^2 -6xy +5y^2.$
典例2
解 (1)①$50.01^2=(50+0.01)^2$
$=50^2 +2×50×0.01 +0.01^2$
$=2500 +1 +0.0001$
$=2501.0001.$
②$49.9^2=(50 -0.1)^2$
$=50^2 -2×50×0.1 +0.1^2$
$=2500 -10 +0.01=2490.01.$
(2)$\because (x+y)^2=16,(x-y)^2=4,$
$\therefore x^2 +2xy + y^2=16,①$
$x^2 -2xy + y^2=4,②$
$①-②,得4xy=12,解得xy=3.$

典例1
解 (1)$(2x+3y)^2$
$=(2x)^2 + 2·(2x)·(3y)+(3y)^2$
$=4x^2 +12xy +9y^2.$
(2)$(\frac{1}{3}m - \frac{1}{2}n)^2$
$=(\frac{1}{3}m)^2 -2·(\frac{1}{3}m)·(\frac{1}{2}n)+(\frac{1}{2}n)^2$
$=\frac{1}{9}m^2 - \frac{1}{3}mn + \frac{1}{4}n^2.$
(3)$(3x - y)^2 - (x - 2y)(x + 2y)$
$=9x^2 -6xy + y^2 -x^2 +4y^2$
$=8x^2 -6xy +5y^2.$
典例2
解 (1)①$50.01^2=(50+0.01)^2$
$=50^2 +2×50×0.01 +0.01^2$
$=2500 +1 +0.0001$
$=2501.0001.$
②$49.9^2=(50 -0.1)^2$
$=50^2 -2×50×0.1 +0.1^2$
$=2500 -10 +0.01=2490.01.$
(2)$\because (x+y)^2=16,(x-y)^2=4,$
$\therefore x^2 +2xy + y^2=16,①$
$x^2 -2xy + y^2=4,②$
$①-②,得4xy=12,解得xy=3.$

1. 运用乘法公式计算$(m-2)^2$的结果是().

A.$m^2 -4$
B.$m^2 -2m +4$
C.$m^2 -4m +4$
D.$m^2 +4m -4$

1.C

2. 运算结果是$x^4y^2 - 2x^2y +1$的是().

A.$(-1 + x^2y^2)^2$
B.$(1 + x^2y^2)^2$
C.$(-1 + x^2y)^2$
D.$(-1 - x^2y)^2$

2.C