典例解析
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)$(3x-y)^2-(x-2y)(x+2y)$
$=9x^2-6xy+y^2-x^2+4y^2$
$=8x^2-6xy+5y^2.$
(2)若$(a+b)^2=13,(a-b)^2=7$,求$a^2+b^2$和$ab$的值.
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) $(3x-y)^2-(x-2y)(x+2y)=9x^2-6xy+y^2-x^2+4y^2=8x^2-6xy+5y^2.$
(4) $\because (a+b)^2=13,(a-b)^2=7$,
$\therefore a^2+2ab+b^2=13,①$
$a^2-2ab+b^2=7,②$
①+②,得$2a^2+2b^2=20$,
$\therefore a^2+b^2=10$.
①-②,得$4ab=6$,
$\therefore ab=\frac{3}{2}$.
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