Cho hệ phương trình $\left\{ \begin{array}{l}{\left( {\dfrac{2}{3}} \right)^{2x - y}} + 6{\left( {\dfrac{2}{3}} \right)^{\dfrac{{2x - y}}{2}}} - 7 = 0\\{3^{{{\log }_9}\left( {x - y} \right)}} = 1\end{array} \right.$. Chọn khẳng định đúng?

$T = \dfrac{{17}}{2}.$

$T = 4.$

$T = \dfrac{{13}}{2}.$

$T = \dfrac{{15}}{2}.$

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

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

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

${x^{\sqrt 3 }} = {y^{\sqrt 3 }}$ 

$y = {x^{ - 4}}$.

$y = {x^4}$.

$y = {x^{ - \dfrac{3}{4}}}$.

$y = \sqrt[3]{x}$.

$x=3+\sqrt{2}$               

$x=\dfrac{-11}{4}$         

$x=3-\sqrt{2}$                 

$x=\dfrac{11}{4}$

$D = \left( { - \infty ;\dfrac{{ - 3 - \sqrt {17} }}{2}} \right] \cup \left[ {\dfrac{{ - 3 + \sqrt {17} }}{2}; + \infty } \right)$

$D = \left( { - \infty ; - 3} \right) \cup \left( {1; + \infty } \right)$

$D = \left( {\dfrac{{ - 3 - \sqrt {17} }}{2}; - 3} \right) \cup \left( {\dfrac{{ - 3 + \sqrt {17} }}{2};1} \right)$            

$D = \left[ {\dfrac{{ - 3 - \sqrt {17} }}{2}; - 3} \right) \cup \left[ {\dfrac{{ - 3 + \sqrt {17} }}{2};1} \right)$