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9. 已知$a$、$b$、$c$均为常数,若$(x - 1)^{2} + bx + c = x^{2} - ax + 16$,则$a + b + c$的值为(
A.18
B.17
C.16
D.15
B
)A.18
B.17
C.16
D.15
答案
9. B 解析:$(x-1)^{2}+bx+c=x^{2}-2x+1+bx+c=x^{2}+(b-$
$2)x+c+1=x^{2}-ax+16,\therefore b-2=-a,c+1=16,\therefore a+b=$
$2,c=15,\therefore a+b+c=2+15=17$.
$2)x+c+1=x^{2}-ax+16,\therefore b-2=-a,c+1=16,\therefore a+b=$
$2,c=15,\therefore a+b+c=2+15=17$.
10. 有两个正方形$A$、$B$,将$A$、$B$并列放置后构造新的图形,分别得到一个长方形(图1)与一个正方形(图2).若图1、图2中阴影部分的面积分别为12与30,则正方形$B$的面积为(
A.3
B.4
C.5
D.6
A
)A.3
B.4
C.5
D.6
答案
10. A 解析:设正方形 A
的边长为 a,正方形 B 的边长为 b. 由题意可知,$a(a+b)-$
$a^{2}-b^{2}=12,(a+b)^{2}-a^{2}-b^{2}=30$,即$ab-b^{2}=12,ab=15,$
$\therefore b^{2}=15-12=3$,即正方形 B 的面积为 3.
的边长为 a,正方形 B 的边长为 b. 由题意可知,$a(a+b)-$
$a^{2}-b^{2}=12,(a+b)^{2}-a^{2}-b^{2}=30$,即$ab-b^{2}=12,ab=15,$
$\therefore b^{2}=15-12=3$,即正方形 B 的面积为 3.
11. 若$x^{2} - 4y^{2} = 5$,则$(x - 2y)^{2}(x + 2y)^{2} =$
25
.答案
11. 25 解析:
$(x-2y)^{2}(x+2y)^{2}=[(x-2y)(x+2y)]^{2}=(x^{2}-4y^{2})^{2}=5^{2}=$
25.
$(x-2y)^{2}(x+2y)^{2}=[(x-2y)(x+2y)]^{2}=(x^{2}-4y^{2})^{2}=5^{2}=$
25.
12. 若正数$m$、$n$满足等式$(m + n - 1)^{2} = (m - 1)^{2} + (n - 1)^{2}$,则$mn =$
$\frac {1}{2}$
.答案
12. $\frac {1}{2}$ 解析:$(m+n-1)^{2}=(m+n)^{2}-2(m+n)+1=$
$m^{2}+n^{2}+2mn-2m-2n+1.(m-1)^{2}+(n-1)^{2}=m^{2}-2m+1+$
$n^{2}-2n+1.\because (m+n-1)^{2}=(m-1)^{2}+(n-1)^{2},\therefore 2mn+1=$
$1+1,\therefore mn=\frac {1}{2}$.
$m^{2}+n^{2}+2mn-2m-2n+1.(m-1)^{2}+(n-1)^{2}=m^{2}-2m+1+$
$n^{2}-2n+1.\because (m+n-1)^{2}=(m-1)^{2}+(n-1)^{2},\therefore 2mn+1=$
$1+1,\therefore mn=\frac {1}{2}$.
13. 计算:
(1)$(-\frac{1}{2}x + 2y)^{2} + (-\frac{1}{2}x - 2y)^{2}$;
(2)$(2m + n - p)(2m - n + p)$;
(3)$(2x + 3y)^{2} - (2x + y)(2x - y)$;
(4)$(2a + b)^{2}(2a - b)^{2}(4a^{2} + b^{2})^{2}$.
(1)$(-\frac{1}{2}x + 2y)^{2} + (-\frac{1}{2}x - 2y)^{2}$;
(2)$(2m + n - p)(2m - n + p)$;
(3)$(2x + 3y)^{2} - (2x + y)(2x - y)$;
(4)$(2a + b)^{2}(2a - b)^{2}(4a^{2} + b^{2})^{2}$.
答案
13. (1)原式$=\frac {1}{4}x^{2}-2xy+4y^{2}+\frac {1}{4}x^{2}+$
$2xy+4y^{2}=\frac {1}{2}x^{2}+8y^{2}$. (2)原式$=[2m+(n-p)][2m-$
$(n-p)]=(2m)^{2}-(n-p)^{2}=4m^{2}-n^{2}+2np-p^{2}$. (3)原
式$=4x^{2}+12xy+9y^{2}-4x^{2}+y^{2}=12xy+10y^{2}$. (4)原式=
$[(2a+b)(2a-b)(4a^{2}+b^{2})]^{2}=[(4a^{2}-b^{2})(4a^{2}+b^{2})]^{2}=$
$(16a^{4}-b^{4})^{2}=256a^{8}-32a^{4}b^{4}+b^{8}$.
$2xy+4y^{2}=\frac {1}{2}x^{2}+8y^{2}$. (2)原式$=[2m+(n-p)][2m-$
$(n-p)]=(2m)^{2}-(n-p)^{2}=4m^{2}-n^{2}+2np-p^{2}$. (3)原
式$=4x^{2}+12xy+9y^{2}-4x^{2}+y^{2}=12xy+10y^{2}$. (4)原式=
$[(2a+b)(2a-b)(4a^{2}+b^{2})]^{2}=[(4a^{2}-b^{2})(4a^{2}+b^{2})]^{2}=$
$(16a^{4}-b^{4})^{2}=256a^{8}-32a^{4}b^{4}+b^{8}$.
14. 阅读下面的材料:
若$x$满足$(9 - x)(x - 4) = 4$,求$(4 - x)^{2} + (x - 9)^{2}$的值.
设$9 - x = a$,$x - 4 = b$,则$(9 - x)(x - 4) = ab = 4$,$a + b = (9 - x) + (x - 4) = 5$,
$\therefore (4 - x)^{2} + (x - 9)^{2} = (9 - x)^{2} + (x - 4)^{2} = a^{2} + b^{2} = (a + b)^{2} - 2ab = 5^{2} - 2×4 = 17$.
请仿照上面的方法解答下列问题:
(1)若$x$满足$(5 - x)(x - 2) = 2$,求$(5 - x)^{2} + (x - 2)^{2}$的值.
(2)如图,已知正方形$ABCD$的边长为$x$,$E$、$F$分别是边$AD$、$DC$上的点,且$AE = 1$,$CF = 3$,长方形$EMFD$的面积是48,分别以$MF$、$DF$为边作正方形.
①$MF =$
②求阴影部分的面积.
]
若$x$满足$(9 - x)(x - 4) = 4$,求$(4 - x)^{2} + (x - 9)^{2}$的值.
设$9 - x = a$,$x - 4 = b$,则$(9 - x)(x - 4) = ab = 4$,$a + b = (9 - x) + (x - 4) = 5$,
$\therefore (4 - x)^{2} + (x - 9)^{2} = (9 - x)^{2} + (x - 4)^{2} = a^{2} + b^{2} = (a + b)^{2} - 2ab = 5^{2} - 2×4 = 17$.
请仿照上面的方法解答下列问题:
(1)若$x$满足$(5 - x)(x - 2) = 2$,求$(5 - x)^{2} + (x - 2)^{2}$的值.
(2)如图,已知正方形$ABCD$的边长为$x$,$E$、$F$分别是边$AD$、$DC$上的点,且$AE = 1$,$CF = 3$,长方形$EMFD$的面积是48,分别以$MF$、$DF$为边作正方形.
①$MF =$
x-1
,$DF =$x-3
.(用含$x$的代数式表示)②求阴影部分的面积.
]答案
14. (1)设$5-x=a,x-$
$2=b$,则$(5-x)(x-2)=ab=2,a+b=(5-x)+(x-2)=3,$
$\therefore (5-x)^{2}+(x-2)^{2}=a^{2}+b^{2}=(a+b)^{2}-2ab=3^{2}-2×2=$
5. (2)①$x-1$ $x-3$ 解析:由题意,得$AD=CD=x,$
$MF=ED,\therefore MF=ED=AD-AE=x-1,DF=CD-CF=$
$x-3$. ②由题意,得$MF· DF=48$,即$(x-1)(x-3)=48.$
由题图可知,阴影部分的面积为$MF^{2}-DF^{2}=(x-1)^{2}-(x-$
$3)^{2}$.设$x-1=a,x-3=b$,则$(x-1)(x-3)=ab=48,a-b=$
$(x-1)-(x-3)=2,\therefore (a+b)^{2}=(a-b)^{2}+4ab=2^{2}+4×$
$48=196,\therefore a+b=\pm 14$.又$\because a＞0,b＞0,\therefore a+b＞0,\therefore a+b=14,\therefore (x-1)^{2}-(x-3)^{2}=a^{2}-b^{2}=(a+b)(a-b)=14×2=28$,即阴影部分的面积是 28.
$2=b$,则$(5-x)(x-2)=ab=2,a+b=(5-x)+(x-2)=3,$
$\therefore (5-x)^{2}+(x-2)^{2}=a^{2}+b^{2}=(a+b)^{2}-2ab=3^{2}-2×2=$
5. (2)①$x-1$ $x-3$ 解析:由题意,得$AD=CD=x,$
$MF=ED,\therefore MF=ED=AD-AE=x-1,DF=CD-CF=$
$x-3$. ②由题意,得$MF· DF=48$,即$(x-1)(x-3)=48.$
由题图可知,阴影部分的面积为$MF^{2}-DF^{2}=(x-1)^{2}-(x-$
$3)^{2}$.设$x-1=a,x-3=b$,则$(x-1)(x-3)=ab=48,a-b=$
$(x-1)-(x-3)=2,\therefore (a+b)^{2}=(a-b)^{2}+4ab=2^{2}+4×$
$48=196,\therefore a+b=\pm 14$.又$\because a＞0,b＞0,\therefore a+b＞0,\therefore a+b=14,\therefore (x-1)^{2}-(x-3)^{2}=a^{2}-b^{2}=(a+b)(a-b)=14×2=28$,即阴影部分的面积是 28.
