a,$\left \{ {{\frac{3}{x-1}+\frac{2}{y+2}=4} \atop {\frac{2}{x-1}+\frac{1}{y+2}=1}} \right.$  (x $\neq$ 1; y $\neq$ -2)
$\left \{ {{\frac{3}{x-1}+\frac{2}{y+2}=4} \atop {\frac{4}{x-1}+\frac{2}{y+2}=2}} \right.$   
$\left \{ {{\frac{1}{x-1}=-2} \atop {2.(-2)+\frac{1}{y+2}=1}} \right.$     
$\left \{ {{\frac{1}{x-1}=-2} \atop {\frac{1}{y+2}=5}} \right.$     
$\left \{ {{x-1=\frac{-1}2} \atop {y+2=\frac{1}5}} \right.$   
$\left \{ {{x=\frac{1}2(TM)} \atop {y=\frac{-9}5(TM)}} \right.$   
vậy...... 
b,$\left \{ {{\frac{3}x+\frac{2}y=5} \atop {\frac{2}x-\frac{3}y=-1}} \right.$ (x;y $\neq$ 0)  
$\left \{ {{\frac{6}x+\frac{4}y=10} \atop {\frac{6}x-\frac{9}y=-3}} \right.$     
$\left \{ {{\frac{13}y=13} \atop {\frac{2}x-\frac{3}y=-1}} \right.$   
$\left \{ {{y=1(TM)} \atop {\frac{2}x-3=-1}} \right.$   
$\left \{ {{y=1} \atop {x=1(TM)}} \right.$   
vậy......  
c,$\left \{ {{\frac{1}x+\frac{1}y=1} \atop {\frac{3}x+\frac{4}y=5}} \right.$   (x;y $\neq$ 0)
$\left \{ {{\frac{3}x+\frac{3}y=3} \atop {\frac{3}x+\frac{4}y=5}} \right.$     
$\left \{ {{\frac{1}y=2} \atop {\frac{1}x+2=1}} \right.$   
$\left \{ {{y=\frac{1}2(TM)} \atop {\frac{1}x=-1}} \right.$ 
$\left \{ {{y=\frac{1}2} \atop {x=-1(TM}} \right.$   
vậy......