Giải thích các bước giải:
$c,\left \{ {{\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}} \atop {\frac{5}{x}-\frac{2}{y}=\frac{8}{3}}} \right.$
Đặt: $a=\frac{1}{x}$ và $b=\frac{1}{y}.$
$⇔\left \{ {{3a+5b=-\frac{3}{2}} \atop {5a-2b=\frac{8}{3}}} \right.$
$⇔\left \{ {{3a+5b=-\frac{3}{2}} \atop { \frac{25}{2}.a-5b=\frac{20}{3}}} \right.$
Lấy vế $(1)+(2):$
$⇒3a+5b+\frac{25}{2}.a-5b = -\frac{3}{2} + \frac{20}{3}$
$⇔\frac{31}{2}.a = \frac{31}{6}$
$⇒a=\frac{1}{3} ; b=-\frac{1}{2}$
$⇒x=3 ; y=-2$
$d,\left \{ {{\frac{3}{5x}+\frac{1}{y}=\frac{1}{10}} \atop {\frac{3}{4x}+\frac{3}{4y}=\frac{1}{12}}} \right.$
Đặt: $a=\frac{1}{x}$ và $b=\frac{1}{y}.$
$⇔\left \{ {{\frac{3}{5}a+b=\frac{1}{10}} \atop {\frac{3}{4}.a+\frac{3}{4}.b=\frac{1}{12}}} \right.$
$⇔\left \{ {{\frac{9}{20}a+\frac{3}{4}.b=\frac{3}{40}} \atop {\frac{3}{4}.a+\frac{3}{4}.b=\frac{1}{12}}} \right.$
Lấy vế $(1)-(2):$
$⇒\frac{9}{20}.a+\frac{3}{4}.b-\frac{3}{4}.a-\frac{3}{4}.b = \frac{3}{40}-\frac{1}{12}$
$⇔\frac{3}{10}.a=\frac{1}{120}$
$⇒a=\frac{1}{36} ; b=\frac{1}{12}$
$⇒x=36 ; y=12$
$e,\left \{ {{\frac{4}{x+2y}-\frac{1}{x-2y}=1} \atop {\frac{20}{x+2y}+\frac{3}{x-2y}=1}} \right.$
Đặt: $a=\frac{1}{x+2y}$ và $b=\frac{1}{x-2y}.$
$⇔\left \{ {{4a-b=1} \atop {20a+3b=1}} \right.$
$⇔\left \{ {{12a-3b=3} \atop {20a+3b=1}} \right.$
Lấy vế $(1)+(2):$
$⇒12a-3b+20a+3b=4$
$⇒32a=4$
$⇒a=\frac{1}{8} = \frac{1}{x+2y}; b=-\frac{1}{2} = \frac{1}{x-2y}$
$⇒\left \{ {{x+2y=8 } \atop {x-2y=-2}} \right.$
Lấy vế $(1)+(2):$
$⇒x+2y+x-2y=8-2$
$⇒x=3 ; y=\frac{5}{2}$
Hôm nay ngẫu hứng thấy anh Cường nên làm cho thôi :>>
