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17. 若关于 $ x $ 的不等式组$\begin{cases}x - m ≤ 0, \\7 - 2x ＜ 1\end{cases}$的整数解共有4个,则 $ m $ 的取值范围是________.
答案
17. $7≤ m ＜8$ 【点拨】本题考查不等式组的整数解,求不等式组的解集,应遵循以下原则:同大取大,同小取小,大小小大中间找,大大小小解不了.
【解析】$\begin{cases} x - m ≤ 0①, \\7 - 2x ＜ 1②, \end{cases}$ 由①得 $x≤ m$,由②得 $x ＞3$,$\therefore$ 解集为 $3 ＜ x≤ m$. $\because$ 不等式组有 4 个整数解 4,5,6,7,$\therefore 7≤ m ＜8$. 故答案为 $7≤ m ＜8$.
【解析】$\begin{cases} x - m ≤ 0①, \\7 - 2x ＜ 1②, \end{cases}$ 由①得 $x≤ m$,由②得 $x ＞3$,$\therefore$ 解集为 $3 ＜ x≤ m$. $\because$ 不等式组有 4 个整数解 4,5,6,7,$\therefore 7≤ m ＜8$. 故答案为 $7≤ m ＜8$.
18. 如图,在四边形纸片ABCD中,AB//CD,将纸片沿EF折叠,点A,D分别落在A',D'处,且A'D'经过点B,FD'交BC于点G,连接EG,EG平分∠BEF.若EG//A'D',∠A + ∠DFE = 150°,则∠GEF的度数是
$30°$
.答案
18. $30°$ 【点拨】本题考查图形的折叠及其性质,角平分线的定义,平行线的性质,准确识图,熟练掌握折叠的性质,平行线的性质是解决问题的关键.
【解析】设 $∠ GEF = α$. $\because EG$ 平分 $∠ BEF$,$\therefore ∠ GEB = ∠ GEF = α$,$\therefore ∠ BEF = 2∠ GEF = 2α$,$\therefore ∠ AEF = 180° - ∠ BEF = 180° -2α$,由折叠的性质得 $∠ A'EF = ∠ AEF = 180° -2α$,$∠ A = ∠ A'$,$\therefore ∠ A'EG = ∠ A'EF - ∠ GEF = 180° -2α -α = 180° -3α$. $\because AB// CD$,$\therefore ∠ DFE = ∠ BEF = 2α$. $\because ∠ A + ∠ DFE = 150°$,$\therefore ∠ A = 150° - ∠ DFE = 150° -2α$,$\therefore ∠ A' = ∠ A = 150° -2α$. $\because EG// A'D'$,$\therefore ∠ A' + ∠ A'EG = 180°$,即 $150° -2α + 180° -3α = 180°$,解得 $α =30°$,$\therefore ∠ GEF = 30°$. 故答案为 $30°$.
【解析】设 $∠ GEF = α$. $\because EG$ 平分 $∠ BEF$,$\therefore ∠ GEB = ∠ GEF = α$,$\therefore ∠ BEF = 2∠ GEF = 2α$,$\therefore ∠ AEF = 180° - ∠ BEF = 180° -2α$,由折叠的性质得 $∠ A'EF = ∠ AEF = 180° -2α$,$∠ A = ∠ A'$,$\therefore ∠ A'EG = ∠ A'EF - ∠ GEF = 180° -2α -α = 180° -3α$. $\because AB// CD$,$\therefore ∠ DFE = ∠ BEF = 2α$. $\because ∠ A + ∠ DFE = 150°$,$\therefore ∠ A = 150° - ∠ DFE = 150° -2α$,$\therefore ∠ A' = ∠ A = 150° -2α$. $\because EG// A'D'$,$\therefore ∠ A' + ∠ A'EG = 180°$,即 $150° -2α + 180° -3α = 180°$,解得 $α =30°$,$\therefore ∠ GEF = 30°$. 故答案为 $30°$.
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答案
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20. (4分)先化简,再求值:$2(x+1)(x-1)+(y-2x)y-(x-y)^2$,其中$x=-1,y=\frac{1}{7}$.
答案
20. 【点拨】本题考查整式的化简与求值,涉及平方差公式、完全平方公式的应用以及合并同类项.
【解析】$2(x+1)(x-1)+(y-2x)y-(x-y)^2$
$=2(x^2 -1)+y^2 -2xy -(x^2 -2xy +y^2)$
$=2x^2 -2 +y^2 -2xy -x^2 +2xy -y^2$
$=x^2 -2$.
当 $x = -1$ 时,$x^2 -2 = (-1)^2 -2 =1 -2 = -1$.
【解析】$2(x+1)(x-1)+(y-2x)y-(x-y)^2$
$=2(x^2 -1)+y^2 -2xy -(x^2 -2xy +y^2)$
$=2x^2 -2 +y^2 -2xy -x^2 +2xy -y^2$
$=x^2 -2$.
当 $x = -1$ 时,$x^2 -2 = (-1)^2 -2 =1 -2 = -1$.
21. (9分)解下列方程组或不等式(组):
(1) $\begin{cases} 2x + y = 5, \\ 3x - 2y = -3; \end{cases}$
(2) $\dfrac{x - 3}{4} ＜ 6 - \dfrac{3 - 4x}{2};$
(3) $\begin{cases} x + 5 ＞ 3, \\ 2x - 1 ≤ x. \end{cases}$
(1) $\begin{cases} 2x + y = 5, \\ 3x - 2y = -3; \end{cases}$
(2) $\dfrac{x - 3}{4} ＜ 6 - \dfrac{3 - 4x}{2};$
(3) $\begin{cases} x + 5 ＞ 3, \\ 2x - 1 ≤ x. \end{cases}$
答案
21. 【点拨】本题考查解一元一次不等式(组)及二元一次方程组.
【解析】(1) $\begin{cases} 2x + y =5,① \\ 3x -2y =-3,② \end{cases}$
① $×2$,得 $4x +2y =10$,③
② $+$ ③,得 $7x =7$,解得 $x =1$,
把 $x =1$ 代入①得 $2 + y =5$,解得 $y =3$,
$\therefore$ 原方程组的解为 $\begin{cases} x =1, \\ y =3. \end{cases}$
(2) $\dfrac{x -3}{4} ＜6 - \dfrac{3 -4x}{2}$
$x -3 ＜24 -2(3 -4x)$
$x -3 ＜24 -6 +8x$
$x -8x ＜24 -6 +3$
$-7x ＜21$
$x ＞-3$.
(3) $\begin{cases} x +5 ＞3,① \\ 2x -1 ≤ x,② \end{cases}$
解不等式①得 $x ＞-2$,解不等式②得 $x≤1$,
$\therefore$ 原不等式组的解集为 $-2 ＜ x≤1$.
【解析】(1) $\begin{cases} 2x + y =5,① \\ 3x -2y =-3,② \end{cases}$
① $×2$,得 $4x +2y =10$,③
② $+$ ③,得 $7x =7$,解得 $x =1$,
把 $x =1$ 代入①得 $2 + y =5$,解得 $y =3$,
$\therefore$ 原方程组的解为 $\begin{cases} x =1, \\ y =3. \end{cases}$
(2) $\dfrac{x -3}{4} ＜6 - \dfrac{3 -4x}{2}$
$x -3 ＜24 -2(3 -4x)$
$x -3 ＜24 -6 +8x$
$x -8x ＜24 -6 +3$
$-7x ＜21$
$x ＞-3$.
(3) $\begin{cases} x +5 ＞3,① \\ 2x -1 ≤ x,② \end{cases}$
解不等式①得 $x ＞-2$,解不等式②得 $x≤1$,
$\therefore$ 原不等式组的解集为 $-2 ＜ x≤1$.
22. (6分)如图,在每个小正方形的边长为1个单位长度的网格中,△ABC的顶点均在格点(网格线的交点)上,直线$ l $经过小正方形的边.
(1)画出线段$ AC $的垂直平分线$ m $;
(2)画出$ △ ABC $关于直线$ l $成轴对称的$ △ A_1B_1C_1 $;
(3)画出$ △ ABC $关于点$ C $成中心对称的$ △ A_2B_2C $.
(1)画出线段$ AC $的垂直平分线$ m $;
(2)画出$ △ ABC $关于直线$ l $成轴对称的$ △ A_1B_1C_1 $;
(3)画出$ △ ABC $关于点$ C $成中心对称的$ △ A_2B_2C $.
答案
22. 【点拨】本题考查网格作图——作线段的垂直平分线,轴对称变换,中心对称变换,熟练掌握线段垂直平分线的性质、中心对称的性质、轴对称的性质是解题的关键.
【解析】(1)如图,直线 $m$ 即为所求.
(2)如图,$△ A_1B_1C_1$ 即为所求.
(3)如图,$△ A_2B_2C$ 即为所求.
