13. 已知$x^2 - 2x = 2$，代数式$(x - 1)^2 + 2022 =$.

13. 2025 【点拨】本题考查完全平方公式,代数式求值.
【解析】$\because x^2-2x=2$,$\therefore (x-1)^2 +2022 =x^2-2x+1+2022=2+1+2022=2025$. 故答案为2025.

$＜/RichMediaCreation＞＜[SEP_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaReference＞＜answer_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD70_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD63_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[SOG_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD55_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PAD_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD59_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD76_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD93_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜seed:tool_call_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PAD_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[eou_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[UNK_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD75_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/answer_never_used_51bce0c785ca2f68081bfa7d91973934＞＜$|FCResponseEnd|$＞＜[PLHD72_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD69_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD89_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD95_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaReference＞＜$|card|＞:＜|superscript|$＞:＜[eou_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[EOGP_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD78_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD56_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD99_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD98_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD43_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD43_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD74_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD63_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD78_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[SOI_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD60_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜answer_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[SOG_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD55_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD74_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD81_never_used_51bce0c785ca2f68081bfa7d91973934]＞$

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15. 若$(x^2 + mx)(x^2 + 2x - n)$的积中不含$x^2$的项与$x^3$的项，则代数式$mn$的值为________.

15. 8 【点拨】本题考查多项式的乘法中不含某项问题.
【解析】$(x^2+mx)(x^2+2x-n)=x^4+2x^3-nx^2+mx^3+2mx^2-mnx=x^4+(m+2)x^3+(2m-n)x^2-mnx$.
$\because (x^2+mx)(x^2+2x-n)$的积中不含$x^2$和$x^3$的项,$\therefore m+2=0,2m-n=0$,$\therefore m=-2,n=-4$,$\therefore mn=-2×(-4)=8$. 故答案为8.

16. 有两个正方形A,B,现将B放在A的内部得图1,将A,B并列放置后构造新的正方形得图2.若图1和图2中阴影部分的面积分别为5和35,则图2的面积为.

16. 75 【点拨】本题考查完全平方公式的应用.
【解析】设正方形A和B的边长分别为$a$和$b$,$\therefore$ 图1阴影部分面积为$(a-b)^2=5$,即$a^2-2ab+b^2=5$,图2阴影部分面积为$b(a+b)+b(a-b)=35$,即$2ab=35$,$\therefore a^2+b^2=40$,$\therefore$ 图2的面积为$(a+b)^2=a^2+2ab+b^2=40+35=75$. 故答案为75.

三、解答题(本大题共9小题,共68分.解答应写出过程)
17. (9分)计算:
(1)$b· b^{3}· b^{5}$;
(2)$(x-y)^{3}(y-x)^{2}$;

(3)$(-2)^{2}+(-2)^{0}+(-\dfrac{1}{2})^{-1}$.

17. 【点拨】本题考查同底数幂的乘法,有理数的混合运算,熟练掌握运算法则是解题的关键.
【解析】(1)$b · b^3 · b^5 = b^{1+3+5}=b^9$.
(2)$(x-y)^3(y-x)^2=(x-y)^3(x-y)^2=(x-y)^{3+2}=(x-y)^5$.
(3)$(-2)^2 + (-2)^0 + (-\frac{1}{2})^{-1}=4+1-2=3$.

18. (16 分)计算:
(1)$(y - \dfrac{1}{2})^2$；
(2)$(2x - y)(3x - 2y) + 4xy$；
(3)$(x + y - 2)(x - y + 2)$；
(4)$(2x - 1)(2x + 1)(4x^2 + 1)$.
·8·

18. 【点拨】本题考查完全平方公式,平方差公式和多项式乘多项式.
【解析】(1)$(y-\frac{1}{2})^2 = y^2 - y + \frac{1}{4}$.
(2)$(2x-y)(3x-2y)+4xy$
$=6x^2-4xy-3xy+2y^2+4xy$
$=6x^2-3xy+2y^2$.
(3)$(x+y-2)(x-y+2)$
$=[x+(y-2)][x-(y-2)]$
$=x^2-(y-2)^2$
$=x^2-(y^2-4y+4)$
$=x^2-y^2+4y-4$.
(4)$(2x-1)(2x+1)(4x^2+1)$
$=(4x^2-1)(4x^2+1)$
$=16x^4-1$.