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7. 如图是一款手推车的平面示意图,其中 $ AB // CD $, $ ∠ 1 = 24° $, $ ∠ 2 = 76° $,则 $ ∠ 3 $ 的度数为（
A.$ 104° $
B.$ 128° $
C.$ 138° $
D.$ 156° $
$ \begin{cases} 6x - 5 ≥ m, \end{cases} $
B
）.A.$ 104° $
B.$ 128° $
C.$ 138° $
D.$ 156° $
$ \begin{cases} 6x - 5 ≥ m, \end{cases} $
答案
7. B 【点拨】本题考查平行线的性质、三角形外角的性质、邻补角的定义.
【解析】如图,
∵ AB // CD,∠1 = 24°,
∴ ∠A = ∠1 = 24°. 又
∵ ∠2 = 76°,∠4 + ∠2 = 180°,
∴ ∠4 = 180° - ∠2 = 104°,
∴ ∠3 = ∠A + ∠4 = 24° + 104° = 128°. 故选 B.
8.若m使得关于x的不等式组$\begin{cases}\\ \dfrac{x}{4} - \dfrac{x - 1}{6} ＜ \dfrac{1}{2}\end{cases}$至少有2个整数解，且关于x,y的方程组$\begin{cases}2x + y = 4, \\ x + 2y = -3m + 2\end{cases}$的解满足$x - y ＞ 10$，则满足条件的整数m有( )个.
A.5
B.4
C.3
D.6
A.5
B.4
C.3
D.6
答案
8. A 【点拨】本题考查解一元一次不等式组和二元一次方程组.
【解析】$\begin{cases} 6x - 5 ≥ m①, \\ \dfrac{x}{4} - \dfrac{x - 1}{6} ＜ \dfrac{1}{2}②, \end{cases}$ 解不等式①得,$x ≥ \dfrac{m + 5}{6}$,解不等式②得,x ＜ 4.
∵ 该不等式组至少有2个整数解,
∴ $\dfrac{m + 5}{6} ≤ 2$,解得m ≤ 7. $\begin{cases} 2x + y = 4③, \\ x + 2y = -3m + 2④, \end{cases}$ ③ - ④得,x - y = 3m + 2.
∵ x - y ＞ 10,
∴ 3m + 2 ＞ 10,解得m ＞ $\dfrac{8}{3}$,
∴ $\dfrac{8}{3}$ ＜ m ≤ 7,
∴ 满足条件的整数m有3,4,5,6,7,共5个. 故选 A.
【解析】$\begin{cases} 6x - 5 ≥ m①, \\ \dfrac{x}{4} - \dfrac{x - 1}{6} ＜ \dfrac{1}{2}②, \end{cases}$ 解不等式①得,$x ≥ \dfrac{m + 5}{6}$,解不等式②得,x ＜ 4.
∵ 该不等式组至少有2个整数解,
∴ $\dfrac{m + 5}{6} ≤ 2$,解得m ≤ 7. $\begin{cases} 2x + y = 4③, \\ x + 2y = -3m + 2④, \end{cases}$ ③ - ④得,x - y = 3m + 2.
∵ x - y ＞ 10,
∴ 3m + 2 ＞ 10,解得m ＞ $\dfrac{8}{3}$,
∴ $\dfrac{8}{3}$ ＜ m ≤ 7,
∴ 满足条件的整数m有3,4,5,6,7,共5个. 故选 A.
二、填空题(本大题共8小题,每小题3分,共24分)
9. 一个多边形的内角和是$1080°$,则这个多边形的边数是
9. 一个多边形的内角和是$1080°$,则这个多边形的边数是
8
.答案
9. 8 【点拨】本题考查多边形的内角和公式.
【解析】设这个多边形的边数是n,则(n - 2) × 180° = 1 080°,解得n = 8. 故答案为8.
【解析】设这个多边形的边数是n,则(n - 2) × 180° = 1 080°,解得n = 8. 故答案为8.
10. 一次知识竞赛共有20道选择题,答对一题得5分;答错或不答,每题扣1分.要使总得分不少于88分,则至少要答对几道题? 若设答对$ x $道题,可列出的不等式为$\underline{\hspace{5cm}}$.
答案
10. $5x - (20 - x) ≥ 88$ 【点拨】本题考查一元一次不等式的应用.
【解析】设答对x道题,则答错或不答(20 - x)道题. 根据题意得,$5x - (20 - x) ≥ 88$. 故答案为$5x - (20 - x) ≥ 88$.
【解析】设答对x道题,则答错或不答(20 - x)道题. 根据题意得,$5x - (20 - x) ≥ 88$. 故答案为$5x - (20 - x) ≥ 88$.
11. 命题“对顶角相等”的逆命题是
相等的角是对顶角
.答案
11. 相等的角是对顶角 【点拨】本题考查逆命题的构造.
【解析】“对顶角相等”的逆命题是“相等的角是对顶角”. 故答案为相等的角是对顶角.
【解析】“对顶角相等”的逆命题是“相等的角是对顶角”. 故答案为相等的角是对顶角.
12. 若$\begin{cases} x=1, \\ y=-2 \end{cases}$是关于$x,y$的二元一次方程$mx+ny=4$的一组解,则$2m-4n-10$的值为________.
答案
12. -2 【点拨】本题考查二元一次方程的解,用整体代换法求代数式的值.
【解析】
∵ $\begin{cases} x = 1, \\ y = -2 \end{cases}$是关于x,y的二元一次方程mx + ny = 4的一组解,
∴ m - 2n = 4,
∴ 2m - 4n - 10 = 2(m - 2n) - 10 = 2 × 4 - 10 = -2. 故答案为-2.
【解析】
∵ $\begin{cases} x = 1, \\ y = -2 \end{cases}$是关于x,y的二元一次方程mx + ny = 4的一组解,
∴ m - 2n = 4,
∴ 2m - 4n - 10 = 2(m - 2n) - 10 = 2 × 4 - 10 = -2. 故答案为-2.
13. 若关于$ x $的不等式组$\begin{cases}2(x + 1) ＞ 4, \\ x ＞ a\end{cases}$的解集是$ x ＞ 1 $，则$ a $的取值范围是________。
答案
13. $a ≤ 1$ 【点拨】本题考查一元一次不等式组的解法,根据一元一次不等式组的解集确定参数的取值范围.
【解析】$\begin{cases} 2(x + 1) ＞ 4①, \\ x ＞ a②, \end{cases}$ 解不等式①得,x ＞ 1.
∵ 原不等式组的解集是x ＞ 1,
∴ a ≤ 1. 故答案为a ≤ 1.
【解析】$\begin{cases} 2(x + 1) ＞ 4①, \\ x ＞ a②, \end{cases}$ 解不等式①得,x ＞ 1.
∵ 原不等式组的解集是x ＞ 1,
∴ a ≤ 1. 故答案为a ≤ 1.
14. 如图,在$△ ABC$中,$AD$是$△ ABC$的角平分线,$E$是$BC$延长线上一点,$∠ EAC = ∠ B$,$∠ E = α$,则$∠ ADE =$ $\boldsymbol{\frac{3}{2}α}$ (用含$α$的式子表示).
答案
14. $90° - \frac{1}{2}α$ 【点拨】本题考查角平分线的定义,三角形内角和定理,三角形外角的性质.
【解析】在△ACE中,∠EAC + ∠ACE + ∠E = 180°.
∵ ∠EAC = ∠B,∠ACE = ∠B + ∠BAC,
∴ ∠B + ∠B + ∠BAC + ∠E = 180°.又
∵ ∠E = α,AD是△ABC的角平分线,
∴ ∠BAC = 2∠BAD,
∴ 2∠B + 2∠BAD + α = 180°,
∴ ∠B + ∠BAD = $\frac{180° - α}{2}$ = 90° - $\frac{1}{2}α$,
∴ ∠ADE = ∠B + ∠BAD = 90° - $\frac{1}{2}α$. 故答案为$90° - \frac{1}{2}α$.
【解析】在△ACE中,∠EAC + ∠ACE + ∠E = 180°.
∵ ∠EAC = ∠B,∠ACE = ∠B + ∠BAC,
∴ ∠B + ∠B + ∠BAC + ∠E = 180°.又
∵ ∠E = α,AD是△ABC的角平分线,
∴ ∠BAC = 2∠BAD,
∴ 2∠B + 2∠BAD + α = 180°,
∴ ∠B + ∠BAD = $\frac{180° - α}{2}$ = 90° - $\frac{1}{2}α$,
∴ ∠ADE = ∠B + ∠BAD = 90° - $\frac{1}{2}α$. 故答案为$90° - \frac{1}{2}α$.
15. 如果关于 $ x $ 的不等式组 $\begin{cases}2 \\x ＜ m + 1\end{cases}$，恰有4个整数解，则 $ m $ 的取值范围是________。
答案
15. $1 ＜ m ≤ 2$ 【点拨】本题考查一元一次不等式组的解法,根据一元一次不等式组整数解的个数确定参数的取值范围.
【解析】$\begin{cases} 3 - \dfrac{3x + 2}{2} ≤ 4①, \\ x ＜ m + 1②, \end{cases}$ 解不等式①得,x ≥ -$\frac{4}{3}$,
∴ 该不等式组的解集为-$\frac{4}{3}$ ≤ x ＜ m + 1.
∵ 该不等式组恰有4个整数解,
∴ 2 ＜ m + 1 ≤ 3,解得1 ＜ m ≤ 2,
∴ m的取值范围为1 ＜ m ≤ 2. 故答案为1 ＜ m ≤ 2.
【解析】$\begin{cases} 3 - \dfrac{3x + 2}{2} ≤ 4①, \\ x ＜ m + 1②, \end{cases}$ 解不等式①得,x ≥ -$\frac{4}{3}$,
∴ 该不等式组的解集为-$\frac{4}{3}$ ≤ x ＜ m + 1.
∵ 该不等式组恰有4个整数解,
∴ 2 ＜ m + 1 ≤ 3,解得1 ＜ m ≤ 2,
∴ m的取值范围为1 ＜ m ≤ 2. 故答案为1 ＜ m ≤ 2.
16. 如图, 在 $△ ABC$ 中, $AG = BG$, $BD = DE = EC$, $CF = 4AF$. 若 $△ ABC$ 的面积为 $45$, 则四边形 $DEFG$ 的面积为$\underline{\hspace{5cm}}$.
答案
16. 21 【点拨】本题考查三角形的面积、三角形中线的性质、等高三角形的面积比与对应底边之比的关系,等底三角形面积比与对应高之比的关系.
【解析】如题图,连接 CG,AE.
∵ BD = DE = EC,
∴ BD = $\frac{1}{3}$BC,
∵ AG = BG,
∴ AG = BG = $\frac{1}{2}$AB,
∴ $S_{△BDG}$ = $\frac{1}{3}S_{△BCG}$ = $\frac{1}{3}$ × $\frac{1}{2}S_{△ABC}$ = $\frac{1}{6}S_{△ABC}$. 同理,$S_{△CEF}$ = $\frac{4}{15}S_{△ABC}$,$S_{△AFG}$ = $\frac{1}{10}S_{△ABC}$,
∴ $S_{四边形DEFG}$ = $S_{△ABC}$ - $S_{△BDG}$ - $S_{△CEF}$ - $S_{△AFG}$ = $\frac{7}{15}S_{△ABC}$ = $\frac{7}{15}$ × 45 = 21. 故答案为21.
【解析】如题图,连接 CG,AE.
∵ BD = DE = EC,
∴ BD = $\frac{1}{3}$BC,
∵ AG = BG,
∴ AG = BG = $\frac{1}{2}$AB,
∴ $S_{△BDG}$ = $\frac{1}{3}S_{△BCG}$ = $\frac{1}{3}$ × $\frac{1}{2}S_{△ABC}$ = $\frac{1}{6}S_{△ABC}$. 同理,$S_{△CEF}$ = $\frac{4}{15}S_{△ABC}$,$S_{△AFG}$ = $\frac{1}{10}S_{△ABC}$,
∴ $S_{四边形DEFG}$ = $S_{△ABC}$ - $S_{△BDG}$ - $S_{△CEF}$ - $S_{△AFG}$ = $\frac{7}{15}S_{△ABC}$ = $\frac{7}{15}$ × 45 = 21. 故答案为21.
$＜[PLHD98_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD63_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/doubaothinking_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD89_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaReference＞＜[EOGP_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜RichMediaReference＞＜[PLHD43_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[CLS_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD57_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD69_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜RichMediaCreation＞＜[PLHD54_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaReference＞＜[EOI_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD96_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD43_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[UNK_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD81_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD100_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD70_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[SILENT_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜doubaothinking_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD65_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD96_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[eou_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD78_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜answer_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD77_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[CLS_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD93_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/seed:tool_call_never_used_51bce0c785ca2f68081bfa7d91973934＞＜/function_never_used_51bce0c785ca2f68081bfa7d91973934＞＜/seed:tool_call_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD79_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[BOS_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜doubao_withdraw＞＜[eou_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD81_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜$|FCResponseEnd|$＞＜answer_never_used_51bce0c785ca2f68081bfa7d91973934＞＜doubao_withdraw＞＜[SOGP_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜doubao_withdraw＞＜[PLHD64_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[EOGP_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/parameter_never_used_51bce0c785ca2f68081bfa7d91973934＞＜$|video|$＞:＜[PLHD100_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[EOG_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD92_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[SEP_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD43_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[BOS_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD85_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD76_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD79_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD58_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/think_never_used_51bce0c785ca2f68081bfa7d91973934＞＜RichMediaCreation＞＜[PLHD91_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜RichMediaReference＞＜$|video|$＞:＜[PLHD82_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD79_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD71_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaShow＞＜[PLHD98_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜$|paragraph|$＞:＜[PLHD43_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[SEP_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[CLS_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜seed:tool_call_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PAD_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD78_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD60_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD71_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD56_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaShow＞＜[PLHD83_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[BOS_never_used_51bce0c785ca2f68081bfa7d91973934]＞$$＜[PLHD43_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD89_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD55_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD97_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD54_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD89_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜hiddenthink＞＜$|card|$＞:＜/think_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD66_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD69_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD78_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD87_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD85_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD43_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD87_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaShow＞＜[PLHD83_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD43_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD57_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD91_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/parameter_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD77_never_used_51bce0c785ca2f68081bfa7d91973934]＞$ $＜[PLHD54_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[bou_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜RichMediaShow＞＜[audio_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[botu_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD63_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaShow＞＜[PLHD63_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/doubaothinking_never_used_51bce0c785ca2f68081bfa7d91973934＞＜[PLHD78_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[PLHD83_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜[botu_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaShow＞＜[PLHD80_never_used_51bce0c785ca2f68081bfa7d91973934]＞＜/RichMediaShow＞＜[EOGP_never_used_51bce0c785ca2f68081bfa7d91973934]＞$
答案
17. 【点拨】本题考查解二元一次方程组.
【解析】(1)$\begin{cases} x + 2y = 1①, \\ 3x - 2y = 5②, \end{cases}$
① + ②,得4x = 6,解得x = $\frac{3}{2}$,
将x = $\frac{3}{2}$代入①,得$\frac{3}{2}$ + 2y = 1,解得y = -$\frac{1}{4}$,
∴ 原方程组的解为$\begin{cases} x = \dfrac{3}{2}, \\ y = -\dfrac{1}{4}. \end{cases}$
(2)$\begin{cases} \dfrac{1}{2}x - \dfrac{1}{3}y = 1①, \\ x - 3y = 2②, \end{cases}$
① × 6,得3x - 2y = 6③,
② × 3,得3x - 9y = 6④,
③ - ④,得7y = 0,解得y = 0,
将y = 0代入②,得x = 2,
∴ 原方程组的解为$\begin{cases} x = 2, \\ y = 0. \end{cases}$
【解析】(1)$\begin{cases} x + 2y = 1①, \\ 3x - 2y = 5②, \end{cases}$
① + ②,得4x = 6,解得x = $\frac{3}{2}$,
将x = $\frac{3}{2}$代入①,得$\frac{3}{2}$ + 2y = 1,解得y = -$\frac{1}{4}$,
∴ 原方程组的解为$\begin{cases} x = \dfrac{3}{2}, \\ y = -\dfrac{1}{4}. \end{cases}$
(2)$\begin{cases} \dfrac{1}{2}x - \dfrac{1}{3}y = 1①, \\ x - 3y = 2②, \end{cases}$
① × 6,得3x - 2y = 6③,
② × 3,得3x - 9y = 6④,
③ - ④,得7y = 0,解得y = 0,
将y = 0代入②,得x = 2,
∴ 原方程组的解为$\begin{cases} x = 2, \\ y = 0. \end{cases}$
