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5. (2025长沙)若$(x-1)(x+3)= x^{2}+mx+n$,则$m+n$的值为()
A. -1
B. -2
C. -3
D. 2
A. -1
B. -2
C. -3
D. 2
答案
A
6. 设有边长分别为$a和b(a＞b)$的A类和B类正方形纸片、长为$a宽为b$的C类矩形纸片若干张.如图所示要拼一个边长为$a+b$的正方形,需要1张A类纸片,1张B类纸片和2张C类纸片.若要拼一个长为$3a+b$,宽为$2a+2b$的矩形,则需要C类纸片的张数为____张.
答案
8
7. 如图,在长为$3a+2$,宽为$2b-1$的长方形铁片上,挖去长为$2a+4$,宽为$b$的小长方形铁片,则剩余部分面积是____.
答案
$4ab-3a-2$
8. (教材变式)先化简,再求值:$(a+b)(a^{2}-ab+b^{2})-(a-b)(a^{2}+b^{2})$,其中$a= \frac{1}{2}$,$b= -2$.
答案
解: 原式$=a^{3}+b^{3}-a^{3}-ab^{2}+a^{2}b+b^{3}$
$=2b^{3}-ab^{2}+a^{2}b$.
当$a=\frac{1}{2},b=-2$时,
原式$=2×(-2)^{3}-\frac{1}{2}×(-2)^{2}+(\frac{1}{2})^{2}×(-2)$
$=-16-2-\frac{1}{2}$
$=-18\frac{1}{2}$.
$=2b^{3}-ab^{2}+a^{2}b$.
当$a=\frac{1}{2},b=-2$时,
原式$=2×(-2)^{3}-\frac{1}{2}×(-2)^{2}+(\frac{1}{2})^{2}×(-2)$
$=-16-2-\frac{1}{2}$
$=-18\frac{1}{2}$.
9. 已知$x+y= 3$,$(x+2)(y+2)= 12$.
(1)求$xy$的值; (2)求$(x+5)(y+5)$的值.
(1)求$xy$的值; (2)求$(x+5)(y+5)$的值.
答案
解: (1)$(x+2)(y+2)=xy+2(x+y)+4=12$.
$\because x+y=3$,
$\therefore xy+6+4=12$,
$\therefore xy=2$;
(2) 原式$=xy+5(x+y)+25$
$=2+5×3+25$
$=42$.
$\because x+y=3$,
$\therefore xy+6+4=12$,
$\therefore xy=2$;
(2) 原式$=xy+5(x+y)+25$
$=2+5×3+25$
$=42$.
10. (教材变式)
【探究规律】(1)计算:①(x-1)(x+1)= ____;
$②(x-1)(x^{2}+x+1)= ____;$
$③(x-1)(x^{3}+x^{2}+x+1)= ____;……【$总结规律】(2)填空$:(x-1)(x^{n}+x^{n-1}+…+x+1)= ____;【$运用规律】(3)已知$(x-1)(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1)= -2,$求$x^{21}$的值.
【探究规律】(1)计算:①(x-1)(x+1)= ____;
$②(x-1)(x^{2}+x+1)= ____;$
$③(x-1)(x^{3}+x^{2}+x+1)= ____;……【$总结规律】(2)填空$:(x-1)(x^{n}+x^{n-1}+…+x+1)= ____;【$运用规律】(3)已知$(x-1)(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1)= -2,$求$x^{21}$的值.
答案
解: (1)$x^{2}-1,x^{3}-1,x^{4}-1$;
(2)$x^{n+1}-1$;
(3)$\because (x-1)(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1)=-2$,
$\therefore x^{7}-1=-2$,
$\therefore x^{7}=-1$,
$\therefore x^{21}=(x^{7})^{3}=(-1)^{3}=-1$.
(2)$x^{n+1}-1$;
(3)$\because (x-1)(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1)=-2$,
$\therefore x^{7}-1=-2$,
$\therefore x^{7}=-1$,
$\therefore x^{21}=(x^{7})^{3}=(-1)^{3}=-1$.
