Giải các hệ phương trình sau:
a) $\{\begin{array}{l}{\log }_x(3x+2y)=2\\ {\log }_y(2x+3y)=2\end{array}$
b) $\{\begin{array}{l}{\log }_x(6x+4y)=2\\ {\log }_y(6y+4x)=2\end{array}$
c) $\{\begin{array}{l}{\log }_2\left(1-\frac{x}{y}\right)=2-{\log }_{2}y\\ {\log }_{\sqrt{\frac{3}{2}}}x+{\log }_{\sqrt{\frac{3}{2}}}y=4\end{array}$
d) $\{\begin{array}{l}{\log }_{y}x-{\log }_2{y}^2=1\\ {\log }_{4}x-{\log }_{4}y=1\end{array}$
e) $\{\begin{array}{l}{\log }_2\left(x^2+y^2+6\right)=4\\ {\log }_{3}x+{\log }_{3}y=1\end{array}$
f) $\{\begin{array}{l}{x}^{{\log }_{2}y}+y^{{\log }_{2}x}=16\\ {\log }_{2}x-{\log }_{2}y=2\end{array}$
g) $\{\begin{array}{l}{x}^{{\log }_{3}y}+2.y^{{\log }_{3}x}=27\\ {\log }_{3}y-{\log }_{3}x=1\end{array}$
h) $\{\begin{array}{l}3.x^{{\log }_{2}y}+2.y^{{\log }_{2}x}=10\\ {\log }_4{x}^2+{\log }_{2}y=2\end{array}$
i) $\{\begin{array}{l}{\log }_x(2x+y-2)=2\\ {\log }_y(2y+x-2)=2\end{array}$
k) $\{\begin{array}{l}{\log }_2(xy)=4\\ {\log }_2\left(\frac{x}{y}\right)=2\end{array}$
l) $\{\begin{array}{l}{\lg }^{2}x={\lg }^{2}y+{\lg }^2(xy)\\ {\lg }^2(x-y)+\lg x.\lg y=0\end{array}$
m) $\{\begin{array}{l}{\log }_{y}x+{\log }_{y}x=\frac{5}{2}\\ {\log }_6\left(x^2+y^2\right)=1\end{array}$
n) $\{\begin{array}{l}{\log }_2(x-y)=5-{\log }_2(x+y)\\ \frac{\lg x-\lg 4}{\lg y-\lg 3}=-1\end{array}$
o) $\{\begin{array}{l}\lg \left(x^2+y^2\right)=1+\lg 8\\ \lg (x+y)-\lg (x-y)=\lg 3\end{array}$
p) $\{\begin{array}{l}{\log }_{x}y=2\\ {\log }_{x+1}(y+23)=3\end{array}$
q) $\{\begin{array}{l}{\log }_{xy}\frac{y}{x}-{\log }_y^{2}x=1\\ {\log }_2(y-x)=1\end{array}$