Giả sử $MN$ cắt đường thẳng qua $A$ song song với $BC$ tại $E$.
Ta có:
$\angle DEN=90^{\circ}-\angle IAE=30^{\circ}$
$\angle EDN=\angle KDA=90^{\circ}-\angle KAD=30^{\circ}$
Suy ra: $\Delta DEN$ cân tại $N\Rightarrow ND=NE$
Kẻ $ML\perp AE$.
Vì $\Delta MLE$ vuông tại $L$ và có $\angle LEM=30^{\circ}$
nên $ME=2ML$ hay $NM+ND=2AH$.

2.
Đặt $t=\ln x\Rightarrow x=e^t\Rightarrow dx=e^tdt$
Đổi cận: $x=e\Rightarrow t=1$
                $x=e^2\Rightarrow t=2$
Khi đó:
$I=\int\limits_1^2\left(\dfrac{1}{t^2}-\dfrac{1}{t}\right)e^tdt$
    $=\int\limits_1^2\dfrac{e^tdt}{t^2}-\int\limits_1^2\dfrac{d(e^t)}{t}$
    $=\int\limits_1^2\dfrac{e^tdt}{t^2}-\dfrac{e^t}{t} \left|\begin{array}{l}2\\1\end{array}\right.+\int\limits_1^2e^td(\dfrac{1}{t})$
    $=\int\limits_1^2\dfrac{e^tdt}{t^2}+e-\dfrac{e^2}{2}-\int\limits_1^2\dfrac{e^tdt}{t^2}=e-\dfrac{e^2}{2}$

1.
Đặt $x=-t\Rightarrow dx=-dt$
Đổi cận: $x=-\pi\Rightarrow t=\pi$
                $x=\pi\Rightarrow t=-\pi$
Khi đó:
$I=\int\limits_{\pi}^{-\pi}\dfrac{-\sin^2(-t)dt}{3^{-t}+1}$
    $=\int\limits_{\pi}^{-\pi}\dfrac{-3^t\sin^2tdt}{3^t+1}$
    $=\int\limits_{-\pi}^{\pi}\dfrac{3^x\sin^2xdx}{3^x+1}$
Suy ra:
$2I=\int\limits_{-\pi}^{\pi}\sin^2xdx$
      $=\int\limits_{-\pi}^{\pi}\dfrac{1-\cos2x}{2}dx$
      $=\left(\dfrac{x}{2}-\dfrac{\sin2x}{4}\right)\left|\begin{array}{l}\pi\\-\pi\end{array}\right.=\pi$
$\Rightarrow I=\dfrac{\pi}{2}$

Điều kiện: $x^2+2y+1\ge0$.
Hệ phương trình tương đương với:
      $\left\{\begin{array}{l}(2y-\sqrt{x^2+2y+1})^2=y^2+(x^2+2y+1)-2xy-2y-1\\y(y-x)=3-3y\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}(2y-\sqrt{x^2+2y+1})^2=(x-y)^2&(1)\\xy=y^2+3y-3&(2)\end{array}\right.$
Ta có:
$(1)\Leftrightarrow \left[\begin{array}{l}2y-\sqrt{x^2+2y+1}=x-y\\2y-\sqrt{x^2+2y+1}=y-x\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\sqrt{x^2+2y+1}=3y-x\\\sqrt{x^2+2y+1}=x+y\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}3y-x\ge0\\x^2+2y+1=(3y-x)^2\end{array}\right.\\\left\{\begin{array}{l}x+y\ge0\\x^2+2y+1=(x+y)^2\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}3y-x\ge0\\-9y^2+2y+1=-6xy\end{array}\right.\\\left\{\begin{array}{l}x+y\ge0\\-y^2+2y+1=2xy\end{array}\right.\end{array}\right.$
Kết hợp với $(2)$ và điều kiện ta được: $(x;y)\in\{(1;1);(\dfrac{415}{51};\dfrac{17}{3})\}$

Ta có:
      $a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$
$\Leftrightarrow a+b+c=ab+bc+ca$
$\Leftrightarrow abc-ab-bc-ca+a+b+c-1=0$
$\Leftrightarrow (a-1)(b-1)(c-1)=0$
$\Leftrightarrow \left[\begin{array}{l}a=1\\b=1\\c=1\end{array}\right.\Rightarrow M=0$

9.
     $\mathop {\lim }\limits_{x \to -\infty}\dfrac{\sqrt{x^2-x+5}}{2x-1}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{\dfrac{\sqrt{x^2-x+5}}{|x|}}{\dfrac{2x-1}{|x|}}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{\sqrt{1-\dfrac{1}{x}+\dfrac{5}{x^2}}}{-2-\dfrac{1}{|x|}}=-\dfrac{1}{2}$