b)$\left \{ {{\frac{1}{x-2}+\frac{1}{y-1}=2} \atop {\frac{2}{x-2}-\frac{3}{y-1}=1}} \right.$
ĐKXĐ: x $\neq$ 2; y $\neq$ 1
$\frac{1}{x-2}$=a; $\frac{1}{y-1}$=b
Khi đó ta có hệ phương trình:
$\left \{ {{a+b=2} \atop {2a-3b=1}} \right.$
⇔ $\left \{ {{b=2-a} \atop {2a-3(2-a)=1}} \right.$
⇔$\left \{ {{b=2-a} \atop {2a-6+3a=1}} \right.$
⇔$\left \{ {{b=2-a} \atop {5a=7}} \right.$
⇔$\left \{ {{b=2-a} \atop {a=\frac{7}{5}}} \right.$
⇔$\left \{ {{b=2-\frac{7}{5}} \atop {a=\frac{7}{5}}} \right.$
⇔$\left \{ {{b=\frac{3}{5}} \atop {a=\frac{7}{5}}} \right.$
⇔$\left \{ {{\frac{1}{y-1}=\frac{3}{5}} \atop {\frac{1}{x+2}=\frac{7}{5}}} \right.$
⇔$\left \{ {{3(y-1)=5} \atop {7(x+2)=5}} \right.$
⇔ $\left \{ {{3y-3=5} \atop {7x+14=5}} \right.$
⇔$\left \{ {{3y=8} \atop {7x=-9}} \right.$
⇔$\left \{ {{y=\frac{8}{3}} \atop {x=\frac{-9}{7}}} \right.$
Vậy hệphươn trình đã cho có nghiệm là (x;y)=(-9/7;8/3)
