Do ABC cân tại A \(\Rightarrow B=C=\dfrac{180^0-A}{2}=30^0\)

Áp dụng định lý sin: \(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=2R\)

\(\Rightarrow\left\{{}\begin{matrix}a=2R.sinA=4\sqrt{3}.sin120^0=6\\b=c=2R.sinB=4\sqrt{3}sin30^0=2\sqrt{3}\\\end{matrix}\right.\)

\(\Rightarrow\) Chu vi: \(a+b+c=6+4\sqrt{3}\)

\(\overrightarrow{AN}=\overrightarrow{AD}+\overrightarrow{DN}\)

\(=\overrightarrow{AC}-\overrightarrow{AB}+\dfrac{1}{2}\cdot\overrightarrow{DC}\)

\(=\overrightarrow{AC}-\overrightarrow{AB}+\dfrac{1}{2}\cdot\overrightarrow{AB}\)

\(=\overrightarrow{AC}-\dfrac{1}{2}\cdot\overrightarrow{AB}\)

=>Chọn D

\(=8\cdot25-72:9=200-8=192\)

= 8 . 25 - 72 : 9

= 200 - 8

= 192

2^3.25-72:3^2

=8.25-72:9

=192

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\) 
          0,4                      0,4          0,4 
\(V_{H_2}=0,4.22,4=8,96l\\ m_{FeCl_2}=0,4.127=50,8g\\ n_{Fe_2O_3}=\dfrac{14}{160}=0,0875\left(mol\right)\\ pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
   \(LTL:\dfrac{0,0875}{1}< \dfrac{0,4}{3}\) 
=> H₂ dư 
\(n_{H_2\left(p\text{ư}\right)}=3n_{Fe_2O_3}=0,2625\left(mol\right)\\ m_{H_2\left(d\right)}=\left(0,4-0,2625\right).2=0,275g\\ n_{Fe}=2n_{Fe_2O_3}=0,175\left(mol\right)\\ m_{Fe}=0,175.56=9,8g\)

Câu 5: 

\(\Leftrightarrow-x^2+7x-9+2x-9=0\)

\(\Leftrightarrow x^2-9x+18=0\)

=>x=3

=>Chọn A

-1/3<-1/2<0<5/12<4/15<1<7/6

\(lim\left(n-\sqrt{n^2-4n+2}\right)\)

\(=lim\dfrac{n^2-\left(n^2-4n+2\right)}{n+\sqrt{n^2-4n+2}}\)

\(=lim\dfrac{4n-2}{n+\sqrt{n^2-4n+2}}\)

\(=lim\dfrac{4-\dfrac{2}{n}}{1+\sqrt{1-\dfrac{4}{n}+\dfrac{2}{n^2}}}\)

\(=2\)