a/$3\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}$ =$3\overrightarrow{OA}+\overrightarrow{OG}+\overrightarrow{GB}+\overrightarrow{OG}+\overrightarrow{GC}+\overrightarrow{OG}+\overrightarrow{GD}$ =$3(\overrightarrow{OA}+\overrightarrow{OG})+(\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD})=0$(đpcm)b/$3MA^2+MB^2+MC^2+MD^2$$=3(\overrightarrow{MO}+\overrightarrow{OA})^2+(\overrightarrow{MO}+\overrightarrow{OB})^2+(\overrightarrow{MO}+\overrightarrow{OC})^2+(\overrightarrow{MO}+\overrightarrow{OD})^2$ $=6MO^2+3OA^2+OB^2+OC^2+OD^2+2\overrightarrow{MO}(3\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD})$ $=6MO^2+3OA^2+OB^2+OC^2+OD^2$ (đpcm)c/ $k^2=3MA^2+MB^2+MC^2+MD^2$$\Leftrightarrow k^2=6MO^2+3OA^2+OB^2+OC^2+OD^2$$\Leftrightarrow k^2=6MO^2+3OA^2+(\overrightarrow{OG}+\overrightarrow{GB})^2+(\overrightarrow{OG}+\overrightarrow{GC})^2+(\overrightarrow{OG}+\overrightarrow{GD})^2$ $\Leftrightarrow k^2=6MO^2+3OA^2+3OG^2+2\overrightarrow{OG}(\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD})$$\Leftrightarrow k^2=6MO^2+3OA^2+3OG^2$ $\begin{cases}\overrightarrow{OA}=\frac{1}{2}\overrightarrow{GA} \\ \overrightarrow{OG}=\frac{1}{2}\overrightarrow{AG} \end{cases}$$\Rightarrow k^2=6MO^2+\frac{3}{2}AG^2$$\Leftrightarrow MO^2=\frac{1}{6}(k^2-\frac{3}{2}AG^2)$

a/$3\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}$ =$3\overrightarrow{OA}+\overrightarrow{OG}+\overrightarrow{GB}+\overrightarrow{OG}+\overrightarrow{GC}+\overrightarrow{OG}+\overrightarrow{GD}$ =$3(\overrightarrow{OA}+\overrightarrow{OG})+(\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD})=0$(đpcm)b/$3MA^2+MB^2+MC^2+MD^2$$=3(\overrightarrow{MO}+\overrightarrow{OA})^2+(\overrightarrow{MO}+\overrightarrow{OB})^2+(\overrightarrow{MO}+\overrightarrow{OC})^2+(\overrightarrow{MO}+\overrightarrow{OD})^2$ $=6MO^2+3OA^2+OB^2+OC^2+OD^2+2\overrightarrow{MO}(3\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD})$ $=6MO^2+3OA^2+OB^2+OC^2+OD^2$ (đpcm)c/ $k^2=3MA^2+MB^2+MC^2+MD^2$$\Leftrightarrow k^2=6MO^2+3OA^2+OB^2+OC^2+OD^2$$\Leftrightarrow k^2=6MO^2+3OA^2+(\overrightarrow{OG}+\overrightarrow{GB})^2+(\overrightarrow{OG}+\overrightarrow{GC})^2+(\overrightarrow{OG}+\overrightarrow{GD})^2$ $\Leftrightarrow k^2=6MO^2+3OA^2+3OG^2+2\overrightarrow{OG}(\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD})$$\Leftrightarrow k^2=6MO^2+3OA^2+3OG^2$ $\begin{cases}\overrightarrow{OA}=\frac{1}{2}\overrightarrow{GA} \\ \overrightarrow{OG}=\frac{1}{2}\overrightarrow{AG} \end{cases}$$\Rightarrow k^2=6MO^2+\frac{3}{2}AG^2$$\Leftrightarrow MO^2=\frac{1}{6}(k^2-\frac{3}{2}AG^2)$ $+k^2<3/2AG^2$ Quỹ tích M thuộc rỗng $+k^2=3/2AG^2$ Quỹ tích M là O$+k^2>3/2AG^2$ Quỹ tích M là đường tròn tâm 0, bán kính R=$\sqrt{\frac{1}{6}(k^2-\frac{3}{2}AG^2)}$

$x^5+y^5=(x+y)(x^4-x^3y-xy^3+x^2y^2+y^4)=[(x^2+y^2)^2-x^2y^2-xy(x^2+y^2)]=(x+y)(1-x^2y^2-xy)$$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)[(x^2+y^2)-xy)=(x+y)(1-xy)$ VT=$|16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|$=$|(x+y)(16-16x^2y^2-16xy-20+20xy+5)|$=$|(x+y)(-16x^2y^2+4xy+1)|$ (1)Theo gt $x^2+y^2=1\Leftrightarrow (x+y)^2-2xy=1\Rightarrow \frac{(x+y)^2-1}{2}=xy$ Đặt t=x+y ($-\sqrt{2}\leq t\leq \sqrt{2}$) (1)=$|t(-16(\frac{t^2-1}{2})^2+4\frac{t^2-1}{2}+1)|$=$|-4t^5+10t^3-5t|$ Xét $f(t)=-4t^5+10t^3-5t$ ($-\sqrt{2}\leq t\leq \sqrt{2}$) $f'(t)=-20t^4+30t^2-5=-5[(2t^2-\frac{3}{2})^2+\frac{11}{4}]<0 \forall t$ $\Rightarrow$hs đồng biến$t -\sqrt{2} 0 \sqrt{2}$$f'(t) | + |$$. | \sqrt{2} | $$. | 0 |$$ f(t) |-\sqrt{2} | $$f(t)\in [-\sqrt{2} ;\sqrt{2}]\Rightarrow |(x+y)(-16x^2y^2+4xy+1)|\in [0;\sqrt{2}]$$ \Rightarrow |16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|\leq \sqrt{2}$(đpcm)

$x^5+y^5=(x+y)(x^4-x^3y-xy^3+x^2y^2+y^4)=[(x^2+y^2)^2-x^2y^2-xy(x^2+y^2)]=(x+y)(1-x^2y^2-xy)$$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)[(x^2+y^2)-xy)=(x+y)(1-xy)$ VT=$|16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|$=$|(x+y)(16-16x^2y^2-16xy-20+20xy+5)|$=$|(x+y)(-16x^2y^2+4xy+1)|$ (1)Theo gt $x^2+y^2=1\Leftrightarrow (x+y)^2-2xy=1\Rightarrow \frac{(x+y)^2-1}{2}=xy$ Đặt t=x+y ($-\sqrt{2}\leq t\leq \sqrt{2}$) (1)=$|t(-16(\frac{t^2-1}{2})^2+4\frac{t^2-1}{2}+1)|$=$|-4t^5+10t^3-5t|$ Xét $f(t)=-4t^5+10t^3-5t$ ($-\sqrt{2}\leq t\leq \sqrt{2}$) $f'(t)=-20t^4+30t^2-5=-5[(2t^2-\frac{3}{2})^2+\frac{11}{4}]<0 \forall t$ $\Rightarrow$hs nghịch biến$t -\sqrt{2} 0 \sqrt{2}$$f'(t) | - |$$. |\sqrt{2} | $$. | 0 |$$ f(t) | -\sqrt{2} | $$f(t)\in [-\sqrt{2} ;\sqrt{2}]\Rightarrow |(x+y)(-16x^2y^2+4xy+1)|\in [0;\sqrt{2}]$$ \Rightarrow |16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|\leq \sqrt{2}$(đpcm)

$x^5+y^5=(x+y)(x^4-x^3y-xy^3+x^2y^2+y^4)=[(x^2+y^2)^2-x^2y^2-xy(x^2+y^2)]=(x+y)(1-x^2y^2-xy)$$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)[(x^2+y^2)-xy)=(x+y)(1-xy)$ VT=$|16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|$=$|(x+y)(16-16x^2y^2-16xy-20+20xy+5)|$=$|(x+y)(-16x^2y^2+4xy+1)|$ (1)Theo gt $x^2+y^2=1\Leftrightarrow (x+y)^2-2xy=1\Rightarrow \frac{(x+y)^2-1}{2}=xy$ Đặt t=x+y ($-\sqrt{2}\leq t\leq \sqrt{2}$) (1)=$|t(-16(\frac{t^2-1}{2})^2+4\frac{t^2-1}{2}+1)|$=$|-4t^5+10t^3-5t|$ Xét $f(t)=-4t^5+10t^3-5t$ ($-\sqrt{2}\leq t\leq \sqrt{2}$) $f'(t)=-20t^4+30t^2-5=-5[(2t^2-\frac{3}{2})^2+\frac{11}{4}]<0 \forall t$ $\Rightarrow$hs nghịch biến$t -\sqrt{2} 0 \sqrt{2}$$f'(t) | \sqrt{2} - |$$. | 0 |$$f(t) | -\sqrt{2}| $$f(t)\in [-\sqrt{2} ;\sqrt{2}]\Rightarrow |(x+y)(-16x^2y^2+4xy+1)|\in [0;\sqrt{2}]$$ \Rightarrow |16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|\leq \sqrt{2}$(đpcm)

$x^5+y^5=(x+y)(x^4-x^3y-xy^3+x^2y^2+y^4)=[(x^2+y^2)^2-x^2y^2-xy(x^2+y^2)]=(x+y)(1-x^2y^2-xy)$$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)[(x^2+y^2)-xy)=(x+y)(1-xy)$ VT=$|16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|$=$|(x+y)(16-16x^2y^2-16xy-20+20xy+5)|$=$|(x+y)(-16x^2y^2+4xy+1)|$ (1)Theo gt $x^2+y^2=1\Leftrightarrow (x+y)^2-2xy=1\Rightarrow \frac{(x+y)^2-1}{2}=xy$ Đặt t=x+y ($-\sqrt{2}\leq t\leq \sqrt{2}$) (1)=$|t(-16(\frac{t^2-1}{2})^2+4\frac{t^2-1}{2}+1)|$=$|-4t^5+10t^3-5t|$ Xét $f(t)=-4t^5+10t^3-5t$ ($-\sqrt{2}\leq t\leq \sqrt{2}$) $f'(t)=-20t^4+30t^2-5=-5[(2t^2-\frac{3}{2})^2+\frac{11}{4}]<0 \forall t$ $\Rightarrow$hs đồng biến$t -\sqrt{2} 0 \sqrt{2}$$f'(t) | + |$$. | \sqrt{2} |$$. | 0 |$$f(t) |-\sqrt{2} | $$f(t)\in [-\sqrt{2} ;\sqrt{2}]\Rightarrow |(x+y)(-16x^2y^2+4xy+1)|\in [0;\sqrt{2}]$$ \Rightarrow |16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|\leq \sqrt{2}$(đpcm)

a/$\begin{cases}SA⊥BC \\ AB⊥BC \end{cases}\Rightarrow BC ⊥(SAB)$$\begin{cases}SA⊥CD \\ AD⊥CD \end{cases}\Rightarrow CD ⊥(SAD)$ $\begin{cases}AC⊥BD \\ SA⊥BD \end{cases}\Rightarrow BD ⊥(SAC)$ b/$\begin{cases}AH⊥SB \\ AH⊥BC \end{cases}\Rightarrow AH ⊥(SBC)\Rightarrow AH⊥ SC$ (1)$\begin{cases}AK⊥SD \\ AK⊥CD \end{cases}\Rightarrow AK ⊥(SCD)\Rightarrow AK ⊥SC$(2)(1)(2)=> SC⊥ (AHK) mặt khác SC ⊥AI =>AH,AK,AI đồng phẳngc/ $\begin{cases}SC⊥HK \\ AC⊥HK \end{cases}\Rightarrow HK⊥ (SAC)$ =>HK ⊥HI

a/$\begin{cases}SA⊥BC \\ AB⊥BC \end{cases}\Rightarrow BC ⊥(SAB)$$\begin{cases}SA⊥CD \\ AD⊥CD \end{cases}\Rightarrow CD ⊥(SAD)$ $\begin{cases}AC⊥BD \\ SA⊥BD \end{cases}\Rightarrow BD ⊥(SAC)$ b/$\begin{cases}AH⊥SB \\ AH⊥BC \end{cases}\Rightarrow AH ⊥(SBC)\Rightarrow AH⊥ SC$ (1)$\begin{cases}AK⊥SD \\ AK⊥CD \end{cases}\Rightarrow AK ⊥(SCD)\Rightarrow AK ⊥SC$(2)(1)(2)=> SC⊥ (AHK) mặt khác SC ⊥AI =>AH,AK,AI đồng phẳngc/ $\begin{cases}SC⊥HK \\ AC⊥HK \end{cases}\Rightarrow HK⊥ (SAC)$ =>HK ⊥HId/$S_ {AHIK}=2S_{AIK}=JK.AI$ với $J=AI\cap HK$$JK=\frac{1}{4} BD=\frac{a\sqrt{2}}{4}$$\frac{1}{AS^2}+\frac{1}{AC^2}=\frac{1}{AI^2}\Rightarrow AI=\sqrt{\frac{6}{5}}a$ => S(AHIK)