Cho biểu thức $A = {3^{ - x + \sqrt x }}$, chọn khẳng định đúng

$a < 0$            

$a > 0$            

$a \in R$         

$a \in Z$ 

 $P = {a^{\frac{1}{2}}}$      

$P = {a^{\frac{9}{2}}}$       

$P = {a^{\frac{{11}}{6}}}$

$P = {a^3}$

${a^n}$ 

${b^n}$          

${a^\alpha }$ 

${b^\alpha }$ 

\(P = {2^{\frac{{181}}{{90}}}}\)      

\(P = {2^{\frac{{181}}{9}}}\)

\(P = {2^{\frac{5}{6}}}\)      

\(P = {2^{\frac{5}{3}}}\)

${\left( {{2^x}} \right)^y} = {2^{x + y}}$ 

$\dfrac{{{2^x}}}{{{2^y}}} = {2^{\frac{x}{y}}}$ 

${2^x}{.2^y} = {2^{x + y}}$ 

${\left( {\dfrac{2}{3}} \right)^x} = \dfrac{{{2^x}}}{{{3^y}}}$ 

\(P = \dfrac{{25}}{{2028}}\)

\(P = 2028\)     

\(P = \dfrac{{{5^3}}}{{{2^{14}}}}\)

\(P = {5^4}{.2^{16}}\)

${2^{\sqrt x }} = {x^{\sqrt 2 }}$       

${3^{\sqrt {xy} }} = {\left( {{3^{\sqrt x }}} \right)^{\sqrt y }}$ 

$\dfrac{{{3^{\sqrt[3]{x}}}}}{{{3^{\sqrt[3]{y}}}}} = {3^{\sqrt[3]{{x - y}}}}$       

${x^{\sqrt 3 }} = {y^{\sqrt 3 }}$