d.
Gọi $I(x;y;z)$ là điểm thỏa mãn: $\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{0}$
Ta có: $\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{0}$
$\Leftrightarrow (x-1;y+1;z-1)+(x+1;y-1;z-3)+(x-3;y-1;z+1)+(x-1;y-3;z-1)=(0;0;0)$
$\Leftrightarrow \left\{\begin{array}{l}4x-4=0\\4y-4=0\\4z-4=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x=1\\y=1\\z=1\end{array}\right.\Leftrightarrow I(1;1;1)$
Khi đó ta có:
     $|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}+\overrightarrow{MD}|$
$=|\overrightarrow{MI}+\overrightarrow{IA}+\overrightarrow{MI}+\overrightarrow{IB}+\overrightarrow{MI}+\overrightarrow{IC}+\overrightarrow{MI}+\overrightarrow{ID}|$
$=|4 \overrightarrow{MI}|=4MI$
Vậy $|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}+\overrightarrow{MD}|$ đạt Min $\Leftrightarrow $ $M$ là hình chiếu của $I$ xuống $(P)$
Đường thẳng $IM$ đi qua $I(1;1;1)$ và có véc-tơ chỉ phương $\overrightarrow{u}=(1;2;-2)$ có phương trình:
                                 $\left\{\begin{array}{l}x=1+t\\y=1+2t\\z=1-2t\end{array}\right.             (t\in\mathbb{R})$
Tọa độ $M$ là nghiệm của hệ: $\left\{\begin{array}{l}x=1+t\\y=1+2t\\z=1-2t\\x+2y-2z+1=0\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}t=\frac{-2}{9}\\x=\frac{7}{9}\\y=\frac{5}{9}\\z=\frac{13}{9}\end{array}\right.$
Vậy: $M(\frac{7}{9};\frac{5}{9};\frac{13}{9})$

c.
Gọi $I(x;y;z)$ là điểm thỏa mãn: $\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}$
Ta có: $\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}$
$\Leftrightarrow (x-1;y+1;z-1)+(x+1;y-1;z-3)+(x-3;y-1;z+1)=(0;0;0)$
$\Leftrightarrow \left\{\begin{array}{l}3x-3=0\\3y-1=0\\3z-3=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x=1\\y=\frac{1}{3}\\z=1\end{array}\right.\Leftrightarrow I(1;\frac{1}{3};1)$
Khi đó ta có:
     $|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}|$
$=|\overrightarrow{MI}+\overrightarrow{IA}+\overrightarrow{MI}+\overrightarrow{IB}+\overrightarrow{MI}+\overrightarrow{IC}|$
$=|3 \overrightarrow{MI}|=3MI$
Vậy $|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}|$ đạt Min $\Leftrightarrow $ $M$ là hình chiếu của $I$ xuống $(P)$
Đường thẳng $IM$ đi qua $I(1;\frac{1}{3};1)$ và có véc-tơ chỉ phương $\overrightarrow{u}=(1;2;-2)$ có phương trình:
                                 $\left\{\begin{array}{l}x=1+t\\y=\frac{1}{3}+2t\\z=1-2t\end{array}\right.             (t\in\mathbb{R})$
Tọa độ $M$ là nghiệm của hệ: $\left\{\begin{array}{l}x=1+t\\y=\frac{1}{3}+2t\\z=1-2t\\x+2y-2z+1=0\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}t=\frac{-2}{27}\\x=\frac{25}{27}\\y=\frac{5}{27}\\z=\frac{31}{27}\end{array}\right.$
Vậy: $M(\frac{25}{27};\frac{5}{27};\frac{31}{27})$

a.
Gọi $I(x;y;z)$ là điểm thỏa mãn: $\overrightarrow{IA}+\overrightarrow{IB}=\overrightarrow{0}$
Ta có: $\overrightarrow{IA}+\overrightarrow{IB}=\overrightarrow{0}$
$\Leftrightarrow (x-1;y+1;z-1)+(x+1;y-1;z-3)=(0;0;0)$
$\Leftrightarrow \left\{\begin{array}{l}2x=0\\2y=0\\2z-4=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x=0\\y=0\\z=2\end{array}\right.\Leftrightarrow I(0;0;2)$
Khi đó ta có:
$|\overrightarrow{MA}+\overrightarrow{MB}|$
$=|\overrightarrow{MI}+\overrightarrow{IA}+\overrightarrow{MI}+\overrightarrow{IB}|$
$=|2 \overrightarrow{MI}|=2MI$
Vậy $|\overrightarrow{MA}+\overrightarrow{MB}|$ đạt Min $\Leftrightarrow $ $M$ là hình chiếu của $I$ xuống $(P)$
Đường thẳng $IM$ đi qua $I(0;0;2)$ và có véc-tơ chỉ phương $\overrightarrow{u}=(1;2;-2)$ có phương trình:
                                 $\left\{\begin{array}{l}x=t\\y=2t\\z=2-2t\end{array}\right.             (t\in\mathbb{R})$
Tọa độ $M$ là nghiệm của hệ: $\left\{\begin{array}{l}x=t\\y=2t\\z=2-2t\\x+2y-2z+1=0\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}t=\frac{1}{3}\\x=\frac{1}{3}\\y=\frac{2}{3}\\z=\frac{4}{3}\end{array}\right.$
Vậy: $M(\frac{1}{3};\frac{2}{3};\frac{4}{3})$

b.
Vì $M\in(d)$ nên tọa độ $M$ có dạng: $M(3b+2;4b-2;b+1)$
Khi đó ta có:
$\overrightarrow{MA}=(3b+1;4b-1;b)$
$\overrightarrow{MB}=(3b+3;4b-3;b-2)$
$\overrightarrow{MC}=(3b-1;4b-3;b+2)$
$\Rightarrow \overrightarrow{MA}+2\overrightarrow{MB}+3\overrightarrow{MC}=(18b+4;24b-16;6b+2)$
$\Rightarrow |\overrightarrow{MA}+2\overrightarrow{MB}+3\overrightarrow{MC}|=\sqrt{(18b+4)^2+(24b-16)^2+(6b+2)^2}$
                                                   $=\sqrt{936b^2-600b+276}$
                                                   $=\sqrt{936(b-\frac{25}{78})^2+\frac{2338}{13}}\ge\sqrt{\frac{2338}{13}}$
Min$|\overrightarrow{MA}+2\overrightarrow{MB}+3\overrightarrow{MC}|=\sqrt{\frac{2338}{13}}\Leftrightarrow b=\frac{25}{78}\Leftrightarrow M(\frac{77}{26};-\frac{28}{39};\frac{103}{78})$

a.
Vì $N\in(d)$ nên tọa độ $N$ có dạng: $N(3a+2;4a-2;a+1)$
Khi đó ta có:
$\overrightarrow{NB}=(3a+3;4a-3;a-2)$
$\overrightarrow{NC}=(3a-1;4a-3;a+2)$
$\Rightarrow \overrightarrow{NB}+\overrightarrow{NC}=(6a+2;8a-6;2a)$
$\Rightarrow |\overrightarrow{NB}+\overrightarrow{NC}|=\sqrt{(6a+2)^2+(8a-6)^2+4a^2}$
                               $=\sqrt{104a^2-72a+40}$
                               $=\sqrt{104(a-\frac{9}{26})^2+\frac{358}{13}}\ge\sqrt{\frac{358}{13}}$
Min$|\overrightarrow{NB}+\overrightarrow{NC}|=\sqrt{\frac{358}{13}}\Leftrightarrow a=\frac{9}{26}\Leftrightarrow N(\frac{79}{26};-\frac{8}{13};\frac{35}{26})$

c.
Theo phần (b) ta có:
      $\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{0}$
$\Leftrightarrow \overrightarrow{MI}+\overrightarrow{IA}+\overrightarrow{MI}+\overrightarrow{IB}+\overrightarrow{MI}+\overrightarrow{IC}+\overrightarrow{MI}+\overrightarrow{ID}=4 \overrightarrow{MI}$
$\Leftrightarrow \overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}+\overrightarrow{MD}=4 \overrightarrow{MI}$

b.
Ta có:
     $\overrightarrow {IA}+\overrightarrow{IB}+\overrightarrow {IC}+\overrightarrow{ID}$
$=\overrightarrow{IE}+\overrightarrow{EA}+\overrightarrow{IE}+\overrightarrow{EB}+\overrightarrow{IF}+\overrightarrow{FC}+\overrightarrow{IF}+\overrightarrow{FD}$
$=2(\overrightarrow{IE}+\overrightarrow {IF})$
$=\overrightarrow {0}$

a.
Ta có:
     $G$ là trọng tâm $\Delta ABC$
$\Leftrightarrow \overrightarrow {GA}+\overrightarrow {GB}+\overrightarrow{GC}=\overrightarrow{0}$
$\Leftrightarrow \overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}=3 \overrightarrow{MG}$
$\Leftrightarrow \overrightarrow{MA}+\overrightarrow {MB}+\overrightarrow{MC}=3 \overrightarrow{MG}$

Cách 2:
Đặt: $2x^2-6x+3=y$, ta được hệ:
      $\left\{\begin{array}{l}2x^2-6x+3=y\\2y^2-6y+3=x\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}2x^2-6x+3=y\\2x^2-6x-2y^2+6y=y-x\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}2x^2-6x+3=y\\(x-y)(2x+2y-5)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}2x^2-6x+3=y\\x=y\end{array}\right.&(I)\\\left\{\begin{array}{l}2x^2-6x+3=y\\2x+2y-5=0\end{array}\right.&(II)\end{array}\right.$
$(I)\Leftrightarrow \left\{\begin{array}{l}x=y\\2x^2-6x+3=x\end{array}\right.$
       $\Leftrightarrow \left[\begin{array}{l}x=y=3\\x=y=\frac{1}{2}\end{array}\right.$
$(II)\Leftrightarrow \left\{\begin{array}{l}2y=5-2x\\4x^2-12x+6=5-2x\end{array}\right.$
         $\Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l}x=\frac{5-\sqrt{21}}{4}\\y=\frac{5+\sqrt{21}}{4}\end{array}\right.\\\left\{\begin{array}{l}x=\frac{5+\sqrt{21}}{4}\\y=\frac{5-\sqrt{21}}{4}\end{array}\right.\end{array}\right.$

Cách 1:
Ta có:
$2(2x^2-6x+3)^2-6(2x^2-6x+3)+3=x$
$\Leftrightarrow 2(4x^2-24x^3+48x^2-36x+9)-12x^2+36x-18+3-x=0$
$\Leftrightarrow 8x^4-48x^3+84x^2-37x+3=0$
$\Leftrightarrow (x-3)(8x^3-24x^2+12x-1)=0$
$\Leftrightarrow (x-3)(2x-1)(4x^2-10x+1)=0$
$\Leftrightarrow \left[\begin{array}{l}x=3\\x=\frac{1}{2}\\x=\frac{5\pm\sqrt{21}}{4}\end{array}\right.$

Điều kiện: $-5\le x\le 13$
Đặt: $\sqrt{x+5}=a;\sqrt{13-x}=b;a;b\ge0$
Ta nhận được hệ:
      $\left\{\begin{array}{l}a-ab+b=-3\\a^2+b^2=18\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=ab-3\\(a+b)^2-2ab=18\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=ab-3\\(ab-3)^2-2ab-18=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=ab-3\\a^2b^2-8ab-9=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=6\\ab=9\end{array}\right.$ (vì $a;b\ge0$)
$\Leftrightarrow \left\{\begin{array}{l}a=3\\b=3\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\sqrt{x+5}=3\\\sqrt{13-x}=3\end{array}\right.\Leftrightarrow x=4$

Ta có:
     $\mathop {\lim }\limits_{x \to 0}\left(\cot x-\frac{1}{\sin x}\right)$
$=\mathop {\lim }\limits_{x \to 0}\frac{\cos x-1}{\sin x}$
$=\mathop {\lim }\limits_{x \to 0}\frac{-2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}$
$=\mathop {\lim }\limits_{x \to 0}\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=0$

2.
Ta có:
     $\int\limits_0^{+\infty}x^3e^{-x^2}dx$
$=\mathop {\lim }\limits_{a \to +\infty}\int\limits_0^ax^3e^{-x^2}dx$
$=\frac{1}{2}\mathop {\lim }\limits_{a \to +\infty}\int\limits_0^a-x^2d(e^{-x^2})$
$=\frac{1}{2}\mathop {\lim }\limits_{a \to +\infty}\left(-x^2e^{-x^2} \left|\begin{array}{l}a\\0\end{array}\right.-\int\limits_0^ae^{-x^2}d(-x^2)
\right)$
$=\frac{1}{2}\mathop {\lim }\limits_{a \to +\infty}\left(-a^2e^{-a^2}-e^{-x^2} \left|\begin{array}{l}a\\0\end{array}\right.\right)$
$=\frac{1}{2}\mathop {\lim }\limits_{a \to +\infty}\left(\frac{-a^2}{e^{a^2}}-\frac{1}{e^{a^2}}+1\right)=\frac{1}{2}$

1.
Ta có:
    $\int\limits_{-\infty}^0xe^xdx$
$=\mathop {\lim }\limits_{a \to -\infty}\int\limits_a^0xd(e^x)$
$=\mathop {\lim }\limits_{a \to -\infty}\left(xe^x \left|\begin{array}{l}0\\a\end{array}\right.-\int\limits_a^0e^xdx\right)$
$=\mathop {\lim }\limits_{a \to -\infty}\left(-ae^a-e^x \left|\begin{array}{l}0\\a\end{array}\right.\right)$
$=\mathop {\lim }\limits_{a \to -\infty}(e^a-ae^a-1)$
$=\mathop {\lim }\limits_{b \to +\infty}(\frac{1}{e^b}+\frac{b}{e^b}-1)=-1$