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一、选择题
1. 为直接应用平方差公式计算，应将 $(a + b - c)(a - b + c)$ 看成(
A.$[(a + b) - c][(a - b) + c]$
B.$[a + (b - c)][a - (b - c)]$
C.$[(a - c) + b][(a + c) - b]$
D.$(a + b - c)[(a - b) + c]$
1. 为直接应用平方差公式计算，应将 $(a + b - c)(a - b + c)$ 看成(
B
)。A.$[(a + b) - c][(a - b) + c]$
B.$[a + (b - c)][a - (b - c)]$
C.$[(a - c) + b][(a + c) - b]$
D.$(a + b - c)[(a - b) + c]$
答案
1. B
2. 下列各式计算正确的是(
A.$(a + b)^2 = a^2 + b^2$
B.$(a - b)^2 = a^2 - b^2$
C.$(2x - y)^2 = 4x^2 - 2xy + y^2$
D.$( \dfrac{1}{2}x + 5 )^2 = \dfrac{1}{4}x^2 + 5x + 25$
D
)。A.$(a + b)^2 = a^2 + b^2$
B.$(a - b)^2 = a^2 - b^2$
C.$(2x - y)^2 = 4x^2 - 2xy + y^2$
D.$( \dfrac{1}{2}x + 5 )^2 = \dfrac{1}{4}x^2 + 5x + 25$
答案
2. D
3. $(x + y)^3$ 等于(
A.$x^3 + y^3$
B.$x^3 + 3xy + y^3$
C.$x^3 + xy^2 + x^2y + y^3$
D.$x^3 + 3xy^2 + 3x^2y + y^3$
D
)。A.$x^3 + y^3$
B.$x^3 + 3xy + y^3$
C.$x^3 + xy^2 + x^2y + y^3$
D.$x^3 + 3xy^2 + 3x^2y + y^3$
答案
3. D
解析
$(x + y)^3 = (x + y)(x + y)^2 = (x + y)(x^2 + 2xy + y^2) = x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3 = x^3 + 3x^2y + 3xy^2 + y^3$，答案选D。
二、填空题
4. $(3x +$$)^2 =$$+ 12x +$$.$
4. $(3x +$$)^2 =$$+ 12x +$$.$
答案
4. 2 $9x^{2}$ 4
5. $(a - b + 2c)^2 =$
$a^{2}+b^{2}+4c^{2}-2ab+4ac-4bc$
。答案
5. $a^{2}+b^{2}+4c^{2}-2ab+4ac-4bc$
6. (1) 若 $x + y = 512$，$x - y = 2$，则代数式 $x^2 - y^2$ 的值是
(2) 如果 $a - b = 3$，$ab = 2$，那么 $a^2 + b^2 =$
1 024
；(2) 如果 $a - b = 3$，$ab = 2$，那么 $a^2 + b^2 =$
13
。答案
6. (1) 1 024 (2) 13
解析
(1) $x^2 - y^2 = (x + y)(x - y) = 512×2 = 1024$
(2) $a^2 + b^2 = (a - b)^2 + 2ab = 3^2 + 2×2 = 9 + 4 = 13$
(2) $a^2 + b^2 = (a - b)^2 + 2ab = 3^2 + 2×2 = 9 + 4 = 13$
三、解答题
7. 解方程组：$\begin{cases} (x - 1)(y - 2) = xy, \\ x(x + 2) - 4y(y - 1) = (x + 2y)(x - 2y). \end{cases}$
7. 解方程组：$\begin{cases} (x - 1)(y - 2) = xy, \\ x(x + 2) - 4y(y - 1) = (x + 2y)(x - 2y). \end{cases}$
答案
7. $\begin{cases}x=\dfrac{4}{3},\\y=-\dfrac{2}{3}\end{cases}$
解析
解：原方程组整理得：
$\begin{cases}2x + y = 2 \\x + 2y = 0\end{cases}$
由第二个方程得 $x = -2y$，代入第一个方程：
$2(-2y) + y = 2$
$-4y + y = 2$
$-3y = 2$
$y = -\dfrac{2}{3}$
则 $x = -2×(-\dfrac{2}{3}) = \dfrac{4}{3}$
$\begin{cases}x = \dfrac{4}{3} \\y = -\dfrac{2}{3}\end{cases}$
$\begin{cases}2x + y = 2 \\x + 2y = 0\end{cases}$
由第二个方程得 $x = -2y$，代入第一个方程：
$2(-2y) + y = 2$
$-4y + y = 2$
$-3y = 2$
$y = -\dfrac{2}{3}$
则 $x = -2×(-\dfrac{2}{3}) = \dfrac{4}{3}$
$\begin{cases}x = \dfrac{4}{3} \\y = -\dfrac{2}{3}\end{cases}$
