登录
8. 在长方形ABCD内,将图1中两张边长分别为a和b(a ＞ b)的正方形纸片按如图2,3所示的两种方式放置(图2,3中两张正方形纸片均有部分重叠),长方形中未被这两张正方形纸片覆盖的部分用阴影表示,设图2中阴影部分的面积为$ S_1 $,图3中阴影部分的面积为$ S_2 $,当$ AD - AB = 2 $时,$ S_2 - S_1 $的值为(
A.$ 2a $
B.$ 2b $
C.$ 2a - 2b $
D.$ -2b $
B
).A.$ 2a $
B.$ 2b $
C.$ 2a - 2b $
D.$ -2b $
答案
B 【点拨】本题考查正方形的性质,整式的混合运算.
【解析】$S_1 = (AB - a) · a + (CD - b)(AD - a) = (AB - a) · a + (AB - b)(AD - a)$,$S_2 = AB(AD - a) + (a - b)(AB - a)$,$\therefore S_2 - S_1 = AB(AD - a) + (a - b)(AB - a) - (AB - a) · a - (AB - b)(AD - a) = (AD - a)(AB - AB + b) + (AB - a)(a - b - a) = b · AD - ab - b · AB + ab = b(AD - AB) = 2b$.故选B.
【解析】$S_1 = (AB - a) · a + (CD - b)(AD - a) = (AB - a) · a + (AB - b)(AD - a)$,$S_2 = AB(AD - a) + (a - b)(AB - a)$,$\therefore S_2 - S_1 = AB(AD - a) + (a - b)(AB - a) - (AB - a) · a - (AB - b)(AD - a) = (AD - a)(AB - AB + b) + (AB - a)(a - b - a) = b · AD - ab - b · AB + ab = b(AD - AB) = 2b$.故选B.
二、填空题(本大题共7小题,每小题3分,共21分)
9. 计算:$a^{2}· a^{4}=$
$(m^{3})^{2}· m^{2}=$
$4a^{2}b· 2b^{2}=$
$(-2a^{2}b)^{3}·(4a^{2}b^{2})=$
$(3x-2y)(2x+2y)=$
9. 计算:$a^{2}· a^{4}=$
$a^6$
;$a^{6}÷ a^{-2}=$$a^8$
;$(m^{3})^{2}· m^{2}=$
$m^8$
;$(-m^{4})^{3}÷ (-m)^{2}=$$-m^{10}$
;$4a^{2}b· 2b^{2}=$
$8a^2b^3$
;$(2a^{4}b^{2}c^{3})÷(4b^{2}c)$$\dfrac{1}{2}a^4c^2$
;$(-2a^{2}b)^{3}·(4a^{2}b^{2})=$
$-32a^8b^5$
;$(a^{2n})^{3}÷(a^{2})^{3n}(n$为整数$)=$$1$
;$(3x-2y)(2x+2y)=$
$6x^2+2xy-4y^2$
;$(2x+3y)^{2}(3y-2x)^{2}=$$81y^4-72x^2y^2+16x^4$
.答案
【点拨】本题考查整式的混合运算,涉及完全平方公式,平方差公式,同底数幂的乘除法,积的乘方与幂的乘方,负整数指数幂法则以及单项式乘单项式法则.
【解析】$a^2 · a^4 = a^6$;$a^6 ÷ a^{-2} = a^{6-(-2)} = a^8$;$(m^3)^2 · m^2 = m^6 · m^2 = m^8$;$(-m^4)^3 ÷ (-m)^2 = (-m^{12}) ÷ m^2 = -m^{10}$;$4a^2b · 2b^2 = 8a^2b^3$;$(2a^4b^2c^3) ÷ (4b^2c) = \dfrac{1}{2}a^4c^2$;$(-2a^2b)^3 · (4a^2b^2) = (-8a^6b^3) · (4a^2b^2) = -32a^8b^5$;$(a^{2n})^3 ÷ (a^2)^{3n} = a^{6n} ÷ a^{6n} = 1$;$(3x-2y)(2x+2y) = 6x^2 +6xy -4xy -4y^2 = 6x^2 +2xy -4y^2$;$(2x+3y)^2(3y-2x)^2 = [(3y+2x)(3y-2x)]^2 = (9y^2 -4x^2)^2 = 81y^4 -72x^2y^2 +16x^4$.故答案为 $a^6;a^8;m^8;-m^{10};8a^2b^3;\dfrac{1}{2}a^4c^2;-32a^8b^5;1;6x^2+2xy-4y^2;81y^4-72x^2y^2+16x^4$.
【解析】$a^2 · a^4 = a^6$;$a^6 ÷ a^{-2} = a^{6-(-2)} = a^8$;$(m^3)^2 · m^2 = m^6 · m^2 = m^8$;$(-m^4)^3 ÷ (-m)^2 = (-m^{12}) ÷ m^2 = -m^{10}$;$4a^2b · 2b^2 = 8a^2b^3$;$(2a^4b^2c^3) ÷ (4b^2c) = \dfrac{1}{2}a^4c^2$;$(-2a^2b)^3 · (4a^2b^2) = (-8a^6b^3) · (4a^2b^2) = -32a^8b^5$;$(a^{2n})^3 ÷ (a^2)^{3n} = a^{6n} ÷ a^{6n} = 1$;$(3x-2y)(2x+2y) = 6x^2 +6xy -4xy -4y^2 = 6x^2 +2xy -4y^2$;$(2x+3y)^2(3y-2x)^2 = [(3y+2x)(3y-2x)]^2 = (9y^2 -4x^2)^2 = 81y^4 -72x^2y^2 +16x^4$.故答案为 $a^6;a^8;m^8;-m^{10};8a^2b^3;\dfrac{1}{2}a^4c^2;-32a^8b^5;1;6x^2+2xy-4y^2;81y^4-72x^2y^2+16x^4$.
10. 用科学记数法表示：$(1.5×10^{-4})^{2}=$
$2.25×10^{-8}$
，$-0.00000311=$$-3.11×10^{-5}$
.答案
【点拨】本题考查科学记数法以及幂的乘方等知识.
【解析】$(1.5×10^{-4})^2 = 2.25×10^{-8}$,$-0.00000311 = -3.11×10^{-5}$.故答案为$2.25×10^{-8}$,$-3.11×10^{-5}$.
【解析】$(1.5×10^{-4})^2 = 2.25×10^{-8}$,$-0.00000311 = -3.11×10^{-5}$.故答案为$2.25×10^{-8}$,$-3.11×10^{-5}$.
11. 比较大小：$-3.2× 10^{-8}$ ______ $-9.7× 10^{-7}$.（填“$＞$”“$＜$”或“$=$”）
答案
＞ 【点拨】本题考查有理数的大小比较.
【解析】$\because |-3.2×10^{-8}| = 3.2×10^{-8} = 0.32×10^{-7}$,$|-9.7×10^{-7}| = 9.7×10^{-7}$,$0.32×10^{-7} ＜ 9.7×10^{-7}$,$\therefore -3.2×10^{-8} ＞ -9.7×10^{-7}$.故答案为$＞$.
【解析】$\because |-3.2×10^{-8}| = 3.2×10^{-8} = 0.32×10^{-7}$,$|-9.7×10^{-7}| = 9.7×10^{-7}$,$0.32×10^{-7} ＜ 9.7×10^{-7}$,$\therefore -3.2×10^{-8} ＞ -9.7×10^{-7}$.故答案为$＞$.
12. 已知 $10^x = 50, 10^y = \frac{1}{2}$，则 $x - y$ 的值为 ______，$9^x ÷ 3^{2y}$ 的值为 ______。
答案
2；81 【点拨】本题考查同底数幂的除法,根据同底数幂的除法运算法则进行计算即可.
【解析】$\because 10^x = 50,10^y = \dfrac{1}{2}$,$\therefore 10^x ÷ 10^y = 10^{x-y}$.又$\because 10^x ÷ 10^y = 50 ÷ \dfrac{1}{2} = 100 = 10^2$,$\therefore x-y=2$,$9^x ÷ 3^{2y} = 3^{2x} ÷ 3^{2y} = 3^{2x-2y} = 3^{2(x-y)} = 3^4 = 81$.故答案为2;81.
【解析】$\because 10^x = 50,10^y = \dfrac{1}{2}$,$\therefore 10^x ÷ 10^y = 10^{x-y}$.又$\because 10^x ÷ 10^y = 50 ÷ \dfrac{1}{2} = 100 = 10^2$,$\therefore x-y=2$,$9^x ÷ 3^{2y} = 3^{2x} ÷ 3^{2y} = 3^{2x-2y} = 3^{2(x-y)} = 3^4 = 81$.故答案为2;81.
13. 若$x$满足$(x+1)^{2x-1}=1$，则$x$的值为________.
答案
0或$\dfrac{1}{2}$ 【点拨】本题考查零指数幂,有理数的乘方.
【解析】当底数为1时,即$x+1=1$时,即$x=0$.把$x=0$代入指数$2x-1$,得$0-1=-1$,则$(x+1)^{2x-1}=1^{-1}=1$,满足条件. 当底数为$-1$时,即$x+1=-1$时,即$x=-2$,把$x=-2$代入指数$2x-1$,得$(-2)×2-1=-5$,则$(x+1)^{2x-1}=(-1)^{-5}=-1$,不满足条件. 当指数为0时,即$2x-1=0$时,即$x=\dfrac{1}{2}$,把$x=\dfrac{1}{2}$代入底数$x+1$,得$\dfrac{1}{2}+1=\dfrac{3}{2}$,则$(x+1)^{2x-1}=(\dfrac{3}{2})^0=1$,满足条件. 综上,$x$的值为0或$\dfrac{1}{2}$.故答案为0或$\dfrac{1}{2}$.
【解析】当底数为1时,即$x+1=1$时,即$x=0$.把$x=0$代入指数$2x-1$,得$0-1=-1$,则$(x+1)^{2x-1}=1^{-1}=1$,满足条件. 当底数为$-1$时,即$x+1=-1$时,即$x=-2$,把$x=-2$代入指数$2x-1$,得$(-2)×2-1=-5$,则$(x+1)^{2x-1}=(-1)^{-5}=-1$,不满足条件. 当指数为0时,即$2x-1=0$时,即$x=\dfrac{1}{2}$,把$x=\dfrac{1}{2}$代入底数$x+1$,得$\dfrac{1}{2}+1=\dfrac{3}{2}$,则$(x+1)^{2x-1}=(\dfrac{3}{2})^0=1$,满足条件. 综上,$x$的值为0或$\dfrac{1}{2}$.故答案为0或$\dfrac{1}{2}$.
14. 若$9x^2 + (k + 1)xy + 4y^2$是关于$x,y$的完全平方式,则常数$k$的值是________.
答案
11或$-13$ 【点拨】本题考查完全平方式,熟练掌握完全平方公式是解题的关键.
【解析】$\because 9x^2 + (k+1)xy +4y^2 = (3x\pm2y)^2$,$\therefore k+1 = \pm2×3×2 = \pm12$,$\therefore k=11$或$k=-13$.故答案为11或$-13$.
【解析】$\because 9x^2 + (k+1)xy +4y^2 = (3x\pm2y)^2$,$\therefore k+1 = \pm2×3×2 = \pm12$,$\therefore k=11$或$k=-13$.故答案为11或$-13$.
15. 下列说法中正确的有
①若$a,b,c$满足$a^2 + b^2 + c^2 = 6$,则$(a + b)^2 + (b + c)^2 + (c + a)^2$的最小值为1；
②若$a,b,c$满足$(2024 - a + b - c)(2025 - a + b - c) = 6$,则$(2024 - a + b - c)^2 + (2025 - a + b - c)^2$的值是13；
③关于$x$的多项式$(x^5 - x^3 + 3x^2 - 1)(x^4 - 5x^2 + x - 6)$的展开式中,$x^4$的系数为$-16$；
④若$x,y$满足$x^2 = y + 2,y^2 = x + 2(x≠y)$,则$2x^2 + 3x + y + 2xy$的值为$-1$.
$· 2·$
②④
.(填序号)①若$a,b,c$满足$a^2 + b^2 + c^2 = 6$,则$(a + b)^2 + (b + c)^2 + (c + a)^2$的最小值为1；
②若$a,b,c$满足$(2024 - a + b - c)(2025 - a + b - c) = 6$,则$(2024 - a + b - c)^2 + (2025 - a + b - c)^2$的值是13；
③关于$x$的多项式$(x^5 - x^3 + 3x^2 - 1)(x^4 - 5x^2 + x - 6)$的展开式中,$x^4$的系数为$-16$；
④若$x,y$满足$x^2 = y + 2,y^2 = x + 2(x≠y)$,则$2x^2 + 3x + y + 2xy$的值为$-1$.
$· 2·$
答案
【点拨】本题考查求整式的值,完全平方公式变形运算,整式混合运算等.
【解析】①$(a+b)^2 + (b+c)^2 + (c+a)^2 = 2(a^2 + b^2 + c^2) + 2ab + 2bc + 2ac = 12 + 2ab + 2bc + 2ac$. $\because (a+b+c)^2 \ge 0$,$\therefore a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \ge 0$,$\therefore 2ab + 2bc + 2ac \ge -6$,$\therefore 12 + 2ab + 2bc + 2ac \ge 12 -6$,$\therefore 12 + 2ab + 2bc + 2ac \ge 6$,$\therefore (a+b)^2 + (b+c)^2 + (c+a)^2 \ge 6$,$\therefore (a+b)^2 + (b+c)^2 + (c+a)^2$的最小值为6,故此项错误;②设$m=2024 -a +b -c$,$n=2025 -a +b -c$,$\therefore mn=6$,$m-n=2024 -a +b -c -(2025 -a +b -c) = -1$,$\therefore (2024 -a +b -c)^2 + (2025 -a +b -c)^2 = m^2 +n^2 = (m-n)^2 + 2mn = (-1)^2 + 2×6 =13$,$\therefore (2024 -a +b -c)^2 + (2025 -a +b -c)^2$的值是13,故此项正确;③$-x^3 · x = -x^4$,$3x^2 · (-5x^2) = -15x^4$,$-1 × x^4 = -x^4$,$\therefore -1 -15 -1 = -17$,关于$x$的多项式的展开式中$x^4$的系数为$-17$,故此项错误;④$\because x^2 = y +2$,$y^2 =x +2$,$\therefore x^2 - y^2 = y -x = (x+y)(x-y)$,$y =x^2 -2$,$\therefore (x+y)(x-y) = -(x-y)$,$\therefore (x-y)(x+y+1)=0$.$\because x\ne y$,$\therefore x+y+1=0$,$\therefore y=-x-1$,$\therefore 2x^2 +3x +y +2xy = 2x^2 +3x -x -1 +2x(-x -1) = 2x^2 +3x -x -1 -2x^2 -2x = -1$,故此项正确.故答案为②④.
【解析】①$(a+b)^2 + (b+c)^2 + (c+a)^2 = 2(a^2 + b^2 + c^2) + 2ab + 2bc + 2ac = 12 + 2ab + 2bc + 2ac$. $\because (a+b+c)^2 \ge 0$,$\therefore a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \ge 0$,$\therefore 2ab + 2bc + 2ac \ge -6$,$\therefore 12 + 2ab + 2bc + 2ac \ge 12 -6$,$\therefore 12 + 2ab + 2bc + 2ac \ge 6$,$\therefore (a+b)^2 + (b+c)^2 + (c+a)^2 \ge 6$,$\therefore (a+b)^2 + (b+c)^2 + (c+a)^2$的最小值为6,故此项错误;②设$m=2024 -a +b -c$,$n=2025 -a +b -c$,$\therefore mn=6$,$m-n=2024 -a +b -c -(2025 -a +b -c) = -1$,$\therefore (2024 -a +b -c)^2 + (2025 -a +b -c)^2 = m^2 +n^2 = (m-n)^2 + 2mn = (-1)^2 + 2×6 =13$,$\therefore (2024 -a +b -c)^2 + (2025 -a +b -c)^2$的值是13,故此项正确;③$-x^3 · x = -x^4$,$3x^2 · (-5x^2) = -15x^4$,$-1 × x^4 = -x^4$,$\therefore -1 -15 -1 = -17$,关于$x$的多项式的展开式中$x^4$的系数为$-17$,故此项错误;④$\because x^2 = y +2$,$y^2 =x +2$,$\therefore x^2 - y^2 = y -x = (x+y)(x-y)$,$y =x^2 -2$,$\therefore (x+y)(x-y) = -(x-y)$,$\therefore (x-y)(x+y+1)=0$.$\because x\ne y$,$\therefore x+y+1=0$,$\therefore y=-x-1$,$\therefore 2x^2 +3x +y +2xy = 2x^2 +3x -x -1 +2x(-x -1) = 2x^2 +3x -x -1 -2x^2 -2x = -1$,故此项正确.故答案为②④.
