mARN : lm khuôn cho sự tổng hợp protein 

rARN : liên kết với protein tạo thành tiểu phần ribosome tham gia vào quá trình dịch mã 

rARN : vận chuyển axit amin tới nơi tổng hợp protein 

TGGTTTGGCTCA

\(\overrightarrow{u}.\left(\overrightarrow{u}-\overrightarrow{v}\right)=u^2-\overrightarrow{u}.\overrightarrow{v}=4-u.v.cos\left(u;v\right)=4-1=3\)

\(\Rightarrow cos\left(\overrightarrow{u};\overrightarrow{u}-\overrightarrow{v}\right)=\dfrac{\overrightarrow{u}.\left(\overrightarrow{u}-\overrightarrow{v}\right)}{\left|\overrightarrow{u}\right|.\left|\overrightarrow{u}-\overrightarrow{v}\right|}\)(*) 

Đặt  \(A=\left|\overrightarrow{u}-\overrightarrow{v}\right|\)

\(A^2=\left(\overrightarrow{u}-\overrightarrow{v}\right)^2=u^2+v^2-2\overrightarrow{u}.\overrightarrow{v}=4+1-2.2.cos60^0=3\)

=> A = \(\sqrt{3}\)

Thay vào (*) ta được \(cos\left(\overrightarrow{u};\overrightarrow{u}-\overrightarrow{v}\right)=\dfrac{3}{2\sqrt{3}}=\dfrac{\sqrt{3}}{2}\Rightarrow\left(\overrightarrow{u};\overrightarrow{u}-\overrightarrow{v}\right)=30^0\)

a, \(\left(x-\dfrac{2}{3}\right)^2=\dfrac{5}{6}=\left(\sqrt{\dfrac{5}{6}}\right)^2\)

TH1 : \(x-\dfrac{2}{3}=\sqrt{\dfrac{5}{6}}\Leftrightarrow x=\dfrac{2}{3}+\sqrt{\dfrac{5}{6}}\)

TH2 : \(x-\dfrac{2}{3}=-\sqrt{\dfrac{5}{6}}\Leftrightarrow x=-\sqrt{\dfrac{5}{6}}+\dfrac{2}{3}\)

b, \(\left(\dfrac{3}{4}-x\right)^3=-8=\left(-2\right)^3\Rightarrow\dfrac{3}{4}-x=-2\Leftrightarrow x=\dfrac{3}{4}+2=\dfrac{11}{4}\)

\(y'=f'\left(\left|2-3x\right|+3\right).\left(\dfrac{1}{2\sqrt{\left(2-3x\right)^2}}\right).2\left(2-3x\right)=0\Rightarrow x=\dfrac{2}{3}\)

TH2 : \(\left[{}\begin{matrix}\left|2-3x\right|+3=-2\left(loại\right)\\\left|2-3x\right|+3=2\left(loại\right)\end{matrix}\right.\)

=> hs có 1 cực trị 

a, \(\left(\dfrac{3}{4}\right)^5x=\left(\dfrac{3}{4}\right)^7\Rightarrow x=\left(\dfrac{3}{4}\right)^7:\left(\dfrac{3}{4}\right)^5=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\)

b, \(\left(x-1\right)^3=27=3^3\Rightarrow x-1=3\Leftrightarrow x=4\)

c, \(\dfrac{64}{4^{x+1}}=4\Leftrightarrow\dfrac{4^3}{4^{x+1}}=4\Leftrightarrow4^{3-\left(x+1\right)}=4\Leftrightarrow4^{2-x}=4\Rightarrow2-x=1\Leftrightarrow x=1\)

d, \(\left(-2\right)^x=4^5.16^2=2^{10}.2^{2^8}=2^{18}=\left(-2\right)^{18}\Rightarrow x=18\)

e, \(\left(x-1\right)^3=-125=\left(-5\right)^3\Rightarrow x-1=-5\Leftrightarrow x=-4\)

f, \(\left(\dfrac{9}{5}-x\right)^2=\dfrac{16}{25}=\left(\dfrac{4}{5}\right)^2\)

TH1 : \(\dfrac{9}{5}-x=\dfrac{4}{5}\Leftrightarrow x=1\)

TH2 : \(\dfrac{9}{5}-x=-\dfrac{4}{5}\Leftrightarrow x=\dfrac{13}{5}\)

\(g'\left(x\right)=f'\left(\left|x-1\right|-2\right).\left(\dfrac{1}{2\sqrt{\left(x-1\right)^2}}\right).2\left(x-1\right)=0\Leftrightarrow x=1\)

TH2 : \(f'\left(\left|x-1\right|-2\right)=0\Rightarrow\left[{}\begin{matrix}\left|x-1\right|-2=0\\\left|x-1\right|-2=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3;x=-1\\x=5;x=-3\end{matrix}\right.\)

=> hs có 5 cực trị 

\(\left|\dfrac{1}{2}x-1\right|=\dfrac{2-x}{2}\)

TH1 : \(\dfrac{1}{2}x-1=\dfrac{2-x}{2}\Rightarrow x-2=2-x\Leftrightarrow2x=4\Leftrightarrow x=2\)

TH2 : \(\dfrac{1}{2}x-1=\dfrac{x-2}{2}\Rightarrow x-2=x-2\)

=> pt có vô số nghiệm 

\(g'\left(x\right)=f'\left(\left|3-2x\right|+3\right).\left(\dfrac{1}{2\sqrt{\left(3-2x\right)^2}}\right).2\left(3-2x\right)=0\Rightarrow x=\dfrac{3}{2}\)

TH2 : \(f'\left(\left|3-2x\right|+3\right)=0\Rightarrow\left[{}\begin{matrix}\left|3-2x\right|+3=\sqrt{2}\left(loại\right)\\\left|3-2x\right|+3=2\left(loại\right)\end{matrix}\right.\)

=> hs có 1 cực trị 

\(g'\left(x\right)=f'\left(\left|x-1\right|+1\right).\left(\dfrac{1}{2\sqrt{\left(x-1\right)^2}}\right).2\left(x-1\right)=0\Rightarrow x=1\)

TH2 : \(\Rightarrow f'\left(\left|x-1\right|+1\right)=0\Rightarrow\left[{}\begin{matrix}\left|x-1\right|+1=1\\\left|x-1\right|+1=2\\\left|x-1\right|+1=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2;x=0\\x=3;x=-1\end{matrix}\right.\)

=> hs có 5 cực trị