7. 如图,正方形ABCD的边长为x,其中$AI=5$,$JC=3$,两个阴影部分都是正方形且面积和为60,则重叠部分四边形FJDI的面积为().

A.28
B.29
C.30
D.31

7. A 【点拨】本题考查正方形和长方形的性质及面积公式,完全平方公式的灵活运用.
【解析】如图,$\because AI=5,JC=3$,两个阴影部分DIEN和DJGM都是正方形,
$\therefore DI = DN = x-5, DJ = DM = x-3$,
$\therefore S_{正方形DIEN} + S_{正方形DJGM} = DI^2 + DJ^2 = 60$,即$(x-5)^2 + (x-3)^2 = 60$①. 又$\because (x-5)-(x-3) = -2, \therefore [(x-5)-(x-3)]^2 = 4$,即$(x-5)^2 + (x-3)^2 - 2(x-5)(x-3) = 4$②,①-②得,$2(x-5)(x-3) = 56, \therefore (x-5)(x-3) = 28, \therefore$ 重叠部分的面积$S_{四边形FJDI} = DI · DJ = (x-5)(x-3) = 28$. 故选 A.

$＜[PLHD55_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD91_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[EOI_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜RichMediaCreation＞＜$|FCResponseBegin|$＞＜[PLHD97_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD63_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜parameter_never_used_51bce0c785ca2f68081bfa7d91973934=＜reflection_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[SOI_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD92_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD86_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD61_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD57_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜doubaothinking_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD60_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜RichMediaShow＞＜[PLHD83_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[SOI_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜$|video|＞:$＜[PLHD98_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜$|superscript|$＞:＜[SPEAK_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD74_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜parameter_never_used_51bce0c785ca2f68081bfa7d91973934=＜[PLHD57_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PAD_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[audio_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD98_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD80_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/seed:tool_call_never_used_51bce0c785ca2f68081bfa7d91973934＞＜RichMediaCreation＞＜$|video|$＞:＜[PLHD99_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD76_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/hiddenthink＞＜[EOI_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD54_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[botu_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD98_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜escapeShell ＜[PLHD62_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD56_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜$|paragraph|$＞:＜[PLHD98_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD67_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD88_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD94_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/seed:tool_call_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[SILENT_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[SPEAK_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaShow＞＜RichMediaCreation＞＜$|FCResponseEnd|$＞＜[EOGP_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD67_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜RichMediaShow＞＜[eotu_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜hiddenthink＞＜[PLHD68_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD89_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD66_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/doubaothinking_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[SPEAK_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜function_never_used_51bce0c785ca2f68081bfa7d91973934=＜[PLHD69_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD86_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/escapeShell＞＜[/audio_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[EOG_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD98_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[audio_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD68_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD93_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaCreation＞＜RichMediaShow＞＜reflection_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD55_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜$|paragraph|$＞:＜[PLHD77_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD85_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/seed:tool_call_never_used_51bce0c785ca2f68081bfa7d91973934＞$

8. C 【点拨】本题考查二元一次方程组的实际应用.
【解析】根据题意可列方程组$\begin{cases}9x=11y,\\(10y+x)-(8x+y)=13.\end{cases}$ 故选 C.

9. 如果不等式 $ 3x - m ＜ 0 $ 的正整数解为 1,2,3,则 $ m $ 的取值范围是().

A.$ 9 ≤ m ＜ 12 $
B.$ 9 ＜ m ≤ 12 $
C.$ m ＜ 12 $
D.$ m ≥ 9 $

9. B 【点拨】本题考查解一元一次不等式,根据一元一次不等式正整数解的情况确定参数的取值范围.
【解析】$\because 3x - m ＜ 0, \therefore x ＜ \frac{m}{3}$. 又$\because$ 该不等式的正整数解为1,2,3,$\therefore 3 ＜ \frac{m}{3} ≤ 4$,即$9 ＜ m ≤ 12$. 故选 B.

10. 如图,在长方形$ABCD$中,$AB=5$,第1次将长方形$ABCD$沿$AB$的方向向右平移4个单位长度,得到长方形$A_1B_1C_1D_1$,第2次将长方形$A_1B_1C_1D_1$沿$A_1B_1$的方向向右平移4个单位长度,得到长方形$A_2B_2C_2D_2$,……,第$n$次将长方形$A_{n-1}B_{n-1}C_{n-1}D_{n-1}$沿$A_{n-1}B_{n-1}$的方向向右平移4个单位长度,得到长方形$A_nB_nC_nD_n(n＞2)$.若$AB_n$的长度为2029,则$n$的值为().


A.505
B.506
C.2021
D.2025

10. B 【点拨】本题考查图形的平移变换,列代数式,解一元一次方程.
【解析】$\because AB=5$,第1次将长方形$ABCD$沿$AB$的方向向右平移4个单位长度,得到长方形$A_1B_1C_1D_1$,此时$BB_1=4,AB_1=AB+BB_1=5+4$;第2次将长方形$A_1B_1C_1D_1$沿$A_1B_1$的方向向右平移4个单位长度,得到长方形$A_2B_2C_2D_2$,此时$B_1B_2=4,AB_2=AB+BB_1+B_1B_2=5+4+4=5+2×4$;……;第$n$次平移后,$B_{n-1}B_n=4,AB_n=AB+BB_1+B_1B_2+\dots+B_{n-1}B_n=5+n×4=5+4n$.$\because AB_n$的长度为2029,$\therefore 5+4n=2029$,解得$n=506$. 故选 B.

11. 2024年6月4日,嫦娥六号携带由玄武岩磨粉、融化,经高科技拉成直径约为0.000 016 7米的丝线织布制作而成的五星红旗在月球背面冉冉升起,经受恶劣环境也能万年不朽,彰显大国实力.数据0.000 016 7用科学记数法表示为.

11. $1.67×10^{-5}$ 【点拨】本题考查科学记数法.
【解析】$0.000\ 016\ 7 = 1.67×10^{-5}$. 故答案为$1.67×10^{-5}$.

12. 写出一个解为$\begin{cases}x=-2,\\ y=1\end{cases}$的二元一次方程组： ______ .

12. $\begin{cases}x+y=-1,\\x-y=-3,\end{cases}$(答案不唯一) 【点拨】本题考查二元一次方程组的解.
【解析】二元一次方程组$\begin{cases}x+y=-1,\\x-y=-3\end{cases}$的解为$\begin{cases}x=-2,\\y=1.\end{cases}$ 故答案为$\begin{cases}x+y=-1,\\x-y=-3\end{cases}$(答案不唯一).

13. 已知$2x - 7y = 5$，那么用含$x$的代数式表示$y$，则$y =$。
·68·

13. $\frac{2x-5}{7}$ 【点拨】本题考查等式的性质.
【解析】由$2x-7y=5$,解得$y=\frac{2x-5}{7}$. 故答案为$\frac{2x-5}{7}$.