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1. 计算$(x-2)(2+x)$的结果是(
A.$x^{2}-4$
B.$4-x^{2}$
C.$x^{2}+4x+4$
D.$x^{2}-4x+4$
A
)A.$x^{2}-4$
B.$4-x^{2}$
C.$x^{2}+4x+4$
D.$x^{2}-4x+4$
答案
1. A
解析
$(x-2)(2+x)=(x-2)(x+2)=x^{2}-2^{2}=x^{2}-4$，结果为选项A。
2. 下列各式中能用平方差公式计算的是(
A.$(x-y)(x-y)$
B.$(x-y)(x+1)$
C.$(x-y)(x+y)$
D.$(x+y)(x+y)$
C
)A.$(x-y)(x-y)$
B.$(x-y)(x+1)$
C.$(x-y)(x+y)$
D.$(x+y)(x+y)$
答案
2. C
3. 下列各式中,为完全平方式的是(
A.$x^{2}-2x-1$
B.$x^{2}-x+1$
C.$x^{2}-x+\frac{1}{4}$
D.$x^{2}-mx+m^{2}$
C
)A.$x^{2}-2x-1$
B.$x^{2}-x+1$
C.$x^{2}-x+\frac{1}{4}$
D.$x^{2}-mx+m^{2}$
答案
3. C
4. 下列各式中,计算正确的是(
A.$(x-2)(x+2)=x^{2}-2$
B.$(a+b)^{2}=a^{2}-ab+b^{2}$
C.$(a+b)^{2}=a^{2}+b^{2}$
D.$(-3a+2)(-3a-2)=9a^{2}-4$
D
)A.$(x-2)(x+2)=x^{2}-2$
B.$(a+b)^{2}=a^{2}-ab+b^{2}$
C.$(a+b)^{2}=a^{2}+b^{2}$
D.$(-3a+2)(-3a-2)=9a^{2}-4$
答案
4. D 解析:$(x - 2)(x + 2) = x^{2} - 4$，故 A 选项不符合题意；$(a + b)^{2} = a^{2} + 2ab + b^{2}$，故 B、C 选项不符合题意；$(-3a + 2)(-3a - 2) = 9a^{2} - 4$，故 D 选项符合题意.
5. 如图,大正方形与小正方形的面积之差是48,则阴影部分的面积是(
A.12
B.18
C.24
D.30
C
)A.12
B.18
C.24
D.30
答案
5. C 解析:设大正方形的边长为$a$，小正方形的边长为$b$，则$AB = BC = a$，$BE = BD = b$，$AE = a - b$，
∵大正方形与小正方形的面积之差是 48，$\therefore a^{2} - b^{2} = 48$，$\therefore S_{△ AEC} = \frac{1}{2}AE· BC = \frac{1}{2}(a - b)a$，$S_{△ AED} = \frac{1}{2}AE· BD = \frac{1}{2}(a - b)· b$，
∴阴影部分的面积为$S_{△ AEC} + S_{△ AED} = \frac{1}{2}(a - b)· a + \frac{1}{2}(a - b)· b = \frac{1}{2}(a - b)· (a + b) = \frac{1}{2}(a^{2} - b^{2}) = \frac{1}{2}×48 = 24$.
∵大正方形与小正方形的面积之差是 48，$\therefore a^{2} - b^{2} = 48$，$\therefore S_{△ AEC} = \frac{1}{2}AE· BC = \frac{1}{2}(a - b)a$，$S_{△ AED} = \frac{1}{2}AE· BD = \frac{1}{2}(a - b)· b$，
∴阴影部分的面积为$S_{△ AEC} + S_{△ AED} = \frac{1}{2}(a - b)· a + \frac{1}{2}(a - b)· b = \frac{1}{2}(a - b)· (a + b) = \frac{1}{2}(a^{2} - b^{2}) = \frac{1}{2}×48 = 24$.
二、填空题(每小题4分,共36分)
6. $(3m+2n)(3m-2n)=$
6. $(3m+2n)(3m-2n)=$
$9m^{2} - 4n^{2}$
.答案
6. $9m^{2} - 4n^{2}$
7. $(2x-3)$(
$2x + 3$
)$=4x^{2}-9$.答案
7. $2x + 3$
8. 计算:$(3x-2)^{2}=$
$9x^{2} - 12x + 4$
.答案
8. $9x^{2} - 12x + 4$
解析
$(3x-2)^{2}=(3x)^{2}-2×3x×2+2^{2}=9x^{2}-12x+4$
9. 若$m^{2}=9$,$n^{2}=3$,则$(m+n)(m-n)=$
6
.答案
9. 6 解析:$\because m^{2} = 9$，$n^{2} = 3$，$\therefore (m + n)(m - n) = m^{2} - n^{2} = 6$.
解析
$(m + n)(m - n) = m^{2} - n^{2}$，因为$m^{2} = 9$，$n^{2} = 3$，所以$m^{2} - n^{2} = 9 - 3 = 6$。
10. 若$(x+3y)^{2}=x^{2}+6xy+Ay^{2}$,则$A=$
9
.答案
10. 9 解析:$\because (x + 3y)^{2} = x^{2} + 6xy + 9y^{2}$，$\therefore A = 9$.
11. 计算:$220^{2}-219×221=$
1
.答案
11. 1 解析:$220^{2} - 219×221 = 220^{2} - (220 - 1)(220 + 1) = 220^{2} - (220^{2} - 1) = 220^{2} - 220^{2} + 1 = 1$.
12. 已知$x^{2}-2x-1=0$,则代数式$(x-1)^{2}+2026=$
2028
.答案
12. 2028 解析:$\because x^{2} - 2x - 1 = 0$，$\therefore x^{2} - 2x = 1$，$\therefore (x - 1)^{2} + 2026 = x^{2} - 2x + 1 + 2026 = 1 + 1 + 2026 = 2028$.
13. 已知$a+b=6$,$ab=7$,则$(a-b)^{2}=$
8
.答案
13. 8 解析:$\because a + b = 6$，$ab = 7$，$\therefore (a - b)^{2} = (a + b)^{2} - 4ab = 6^{2} - 4×7 = 8$.
解析
$(a - b)^{2} = (a + b)^{2} - 4ab$，因为$a + b = 6$，$ab = 7$，所以$(a - b)^{2} = 6^{2} - 4×7 = 36 - 28 = 8$。
14. 观察以下等式:
$(x+2y)^{2}+(2x-y)^{2}=5(x^{2}+y^{2})$;
$(2x+3y)^{2}+(3x-2y)^{2}=13(x^{2}+y^{2})$;
$(3x+4y)^{2}+(4x-3y)^{2}=25(x^{2}+y^{2})$;
$(4x+6y)^{2}+(6x-4y)^{2}=52(x^{2}+y^{2})$.
运用你所发现的规律解决以下问题:
已知$x$、$y$为实数,$x^{2}+y^{2}=1$,则$(6x+8y)^{2}$的最大值为
$(x+2y)^{2}+(2x-y)^{2}=5(x^{2}+y^{2})$;
$(2x+3y)^{2}+(3x-2y)^{2}=13(x^{2}+y^{2})$;
$(3x+4y)^{2}+(4x-3y)^{2}=25(x^{2}+y^{2})$;
$(4x+6y)^{2}+(6x-4y)^{2}=52(x^{2}+y^{2})$.
运用你所发现的规律解决以下问题:
已知$x$、$y$为实数,$x^{2}+y^{2}=1$,则$(6x+8y)^{2}$的最大值为
100
.答案
14. 100 解析:由等式的规律可知，$(6x + 8y)^{2} + (8x - 6y)^{2} = (6^{2} + 8^{2})(x^{2} + y^{2}) = 100(x^{2} + y^{2})$，$\therefore (6x + 8y)^{2} = 100(x^{2} + y^{2}) - (8x - 6y)^{2}$，$\because x^{2} + y^{2} = 1$，$\therefore (6x + 8y)^{2} = 100 - (8x - 6y)^{2}$，$\because (8x - 6y)^{2} ≥ 0$，$\therefore 0 ≤ 100 - (8x - 6y)^{2} ≤ 100$，$\therefore (6x + 8y)^{2}$的最大值为 100.
