Đáp án:
g) $\left \{ {{x=5} \atop {y=-2}} \right.$
h) $\left \{ {{x=2} \atop {y=5}} \right.$
Giải thích các bước giải:
g) (I) $\left \{ {{\frac{2}{2x-y}+ \frac{3}{x-2y}= \frac{1}{2} } \atop { \frac{2}{2x-y}-\frac{1}{x-2y}=\frac{1}{18} }} \right.$
Đặt $\left \{ {{\frac{1}{2x-y}=a } \atop {\frac{1}{x-2y}=b }} \right.$
(I)⇔ $\left \{ {{2a+3b=\frac{1}{2} } \atop {2a-b=\frac{1}{18} }} \right.$ ⇔ $\left \{ {{4b=\frac{4}{9} } \atop {2a-b= \frac{1}{18}}} \right.$
⇔$\left \{ {{b=\frac{1}{9}} \atop {2a-\frac{1}{9}= \frac{1}{18} }} \right.$ ⇔ $\left \{ {{a=\frac{1}{12} } \atop {b=\frac{1}{9} }} \right.$
⇒ $\left \{ {{\frac{1}{2x-y}= \frac{1}{12} } \atop {\frac{1}{x-2y}=\frac{1}{9}}} \right.$ ⇔$\left \{ {{2x-y=12} \atop {x-2y=9}} \right.$
⇔$\left \{ {{2x-y=12} \atop {2x-4y=18}} \right.$ ⇔ $\left \{ {{3y=-6} \atop {x-2y=9}} \right.$
⇔$\left \{ {{y=-2} \atop {x-2(-2)=9}} \right.$ ⇔$\left \{ {{x=5} \atop {y=-2}} \right.$
Vậy $\left \{ {{x=5} \atop {y=-2}} \right.$
h) $\left \{ {{\frac{x+y}{3}+ \frac{2}{3}=3 } \atop {\frac{4x-y}{6}+\frac{x}{4}=1 }} \right.$ ⇔$\left \{ {{x+y+2=9} \atop {8x-2y+3x=12}} \right.$
⇔$\left \{ {{x+y=7} \atop {11x-2y=12}} \right.$ ⇔$\left \{ {{2x+2y=14} \atop {11x-2y=12}} \right.$
⇔ $\left \{ {{13x=26} \atop {x+y=7}} \right.$ ⇔$\left \{ {{x=2} \atop {y=5}} \right.$
Vậy$\left \{ {{x=2} \atop {y=5}} \right.$
