登录
16.「★★☆」如图，点E,F是正方形ABCD的边AB的三等分点，点P是对角线AC上的动点，当PE+PF取得最小值时，$\frac{AP}{PC}$的值是
$\dfrac{2}{7}$
。答案
答案 $\dfrac{2}{7}$
解析 如图,作点$F$关于$AC$的对称点$F'$,连接$PF'$,则$PF'=PF$,且点$F'$在$AD$上,$\therefore PE+PF=PE+PF'$,连接$EF'$交$AC$于点$P'$,
当点$P$在点$P'$位置时,$PE+PF$取得最小值,
过点$F'$作$F'K⊥ AD$,交$AC$于点$K$,
设正方形$ABCD$的边长为$a$,则$AF'=AF=\dfrac{2}{3}a$,$AE=\dfrac{1}{3}a$,
$\because$ 四边形$ABCD$是正方形,
$\therefore ∠ F'AK=45°$,$∠ P'AE=45°$,$AC=\sqrt{2}a$,
$\because F'K⊥ AF'$,$\therefore ∠ F'AK=∠ F'KA=45°$,
$\therefore AK=\dfrac{2\sqrt{2}}{3}a$,$∠ F'KA=∠ P'AE$,$F'K=AF'=\dfrac{2}{3}a$,
$\because ∠ F'P'K=∠ EP'A$,$\therefore △ F'KP' ∽ △ EAP'$,
$\therefore \dfrac{F'K}{AE}=\dfrac{KP'}{AP'}=2$,$\therefore AP'=\dfrac{1}{3}AK=\dfrac{2\sqrt{2}}{9}a$,
$\therefore CP'=AC-AP'=\dfrac{7\sqrt{2}}{9}a$,$\therefore \dfrac{AP'}{CP'}=\dfrac{2}{7}$,
$\therefore$ 当$PE+PF$取得最小值时,$\dfrac{AP}{PC}$的值是$\dfrac{2}{7}$.
三、解答题(共72分)(答案含评分细则)
17.「2026河南洛阳宜阳期中改编，★☆」(9分)选择合适的方法解下列方程：
(1)$x^2 -6x +1=0$
(2)$x^2 -x=12$
(3)$2x^2 -7x +3=0$
17.「2026河南洛阳宜阳期中改编，★☆」(9分)选择合适的方法解下列方程：
(1)$x^2 -6x +1=0$
(2)$x^2 -x=12$
(3)$2x^2 -7x +3=0$
答案
解析 (1)移项得$x^2-6x=-1$,
配方得$x^2-6x+9=8$,即$(x-3)^2=8$,
直接开平方得$x-3=\pm2\sqrt{2}$,
$\therefore x_1=3+2\sqrt{2}$,$x_2=3-2\sqrt{2}$. $··········································$ (3分)
(2)方程化为一般形式为$x^2-x-12=0$,
$\therefore (x-4)(x+3)=0$,$\therefore x-4=0$或$x+3=0$,
$\therefore x_1=4$,$x_2=-3$. $·$ (6分)
(3)$\because a=2$,$b=-7$,$c=3$,$\therefore \Delta=(-7)^2-4×2×3=25＞0$,$\therefore x=\dfrac{7\pm\sqrt{25}}{2×2}$,$\therefore x_1=3$,$x_2=\dfrac{1}{2}$.$·$ (9分)
配方得$x^2-6x+9=8$,即$(x-3)^2=8$,
直接开平方得$x-3=\pm2\sqrt{2}$,
$\therefore x_1=3+2\sqrt{2}$,$x_2=3-2\sqrt{2}$. $··········································$ (3分)
(2)方程化为一般形式为$x^2-x-12=0$,
$\therefore (x-4)(x+3)=0$,$\therefore x-4=0$或$x+3=0$,
$\therefore x_1=4$,$x_2=-3$. $·$ (6分)
(3)$\because a=2$,$b=-7$,$c=3$,$\therefore \Delta=(-7)^2-4×2×3=25＞0$,$\therefore x=\dfrac{7\pm\sqrt{25}}{2×2}$,$\therefore x_1=3$,$x_2=\dfrac{1}{2}$.$·$ (9分)
