Giải các hệ phương trình sau:
a, $\left\{\begin{matrix} &x(y^{2} - 3) - 2 = x^{2} - y^{2} \\ &log_{4}(x - 1) + log_{4}(2y^{2} - 3) = \frac{1}{2} + log_{2}y \end{matrix}\right.$
b, $\left\{\begin{matrix} &(x^{2} + y^{2})(x + y) + 2xy = x + y \\ &log_{2}\sqrt{x + y} = log_{3}(\sqrt{x^{2} + y^{2} + 1} - 1) \end{matrix}\right.$
c, $\left\{\begin{matrix} &xlog_{2}3 + log_{2}y = y + log_{2}\frac{3x}{2} \\ &xlog_{3}12 + log_{3}x = y + log_{3}\frac{2y}{3} \end{matrix}\right.$
d, $\left\{\begin{matrix} &(3x + 1)\sqrt{9y^{2} + 6y + 2} - y + 1 = 4x\sqrt{16y^{2} + 1} \\ &2012^{x} - 2012^{y} = (log_{3}y - log_{3}x)(12 + 4xy) \end{matrix}\right.$
e, $\left\{\begin{matrix} &1 + \sqrt{x + y + 1} = 4(x + y)^{2} + \sqrt{3(x + y)} \\ &log_{4}(3x + 2y)^{2} + log_{\sqrt{2}}\sqrt{x + 1} = 4 \end{matrix}\right.$
f, $\left\{\begin{matrix} &(2^{x} + \frac{1}{2^{x}})^{y} = (2^{y} + \frac{1}{2^{y}})^{x} \\ &e^{x} + (x^{3} - y)ln(y^{2} + x + 2) = e^{\sqrt[3]{x}} \end{matrix}\right.$
g, $\left\{\begin{matrix} &4\sqrt{1 + x} + xy\sqrt{4 + y^{2}} = 0 \\ &log_{2}x = 2^{y + 2} \end{matrix}\right.$