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Giải thích các bước giải:

ĐKXĐ: $y\ne 0, y\ne -2$

Ta có:

$\begin{cases}|x-10|+\dfrac{|5y-10|}{5y}=12\\ 2|x-10|-\dfrac{|2y-4|}{y+2}=5\end{cases}$

$\to \begin{cases}|x-10|+\dfrac{|y-2|}{y}=12\\ 2|x-10|-\dfrac{|2y-4|}{y+2}=5\end{cases}$

$\to \begin{cases}|x-10|=12-\dfrac{|y-2|}{y}\\ 2|x-10|-2\cdot\dfrac{|y-2|}{y+2}=5\end{cases}$

$\to 2\cdot (12-\dfrac{|y-2|}y-2\cdot\dfrac{|y-2|}{y+2}=5$

$\to 24-2\cdot \dfrac{|y-2|}y-2\cdot\dfrac{|y-2|}{y+2}=5$

$\to 24-2\cdot |y-2|\cdot (\dfrac1y+\dfrac1{y+2})=5$

$\to 2\cdot |y-2|\cdot (\dfrac1y+\dfrac1{y+2})=19$

Trường hợp: $y\ge 2\to y-2\ge 0\to |y-2|=y-2$

$\to 2\cdot (y-2)\cdot (\dfrac1y+\dfrac1{y+2})=19$

$\to \dfrac{2\left(2y+2\right)\left(y-2\right)}{y\left(y+2\right)}=19$

$\to 2\left(2y+2\right)\left(y-2\right)=19y\left(y+2\right)$

$\to y=-\dfrac{21+\sqrt{321}}{15},\:y=-\dfrac{21-\sqrt{321}}{15}$

$\to |x-10|+\dfrac{|-\dfrac{21+\sqrt{321}}{15}-2|}{-\dfrac{21+\sqrt{321}}{15}}=12$

$\to x=-\dfrac{73-\sqrt{321}}{4}+10\quad \mathrm{hoặc}\quad \:x=22-\dfrac{\sqrt{321}-25}{4}$

Hoặc

$\to |x-10|+\dfrac{|-\dfrac{21-\sqrt{321}}{15}-2|}{-\dfrac{21-\sqrt{321}}{15}}=12$

$\to x=-\dfrac{73+\sqrt{321}}{4}+10\quad \mathrm{hoặc}\quad \:x=22+\dfrac{25+\sqrt{321}}{4}$

Trường hợp: $y<2\to y-2<0\to |y-2|=-(y-2)$

$\to  -2\cdot (y-2)\cdot (\dfrac1y+\dfrac1{y+2})=19$

$\to -\dfrac{2\left(2y+2\right)\left(y-2\right)}{y\left(y+2\right)}=19$

$\to -2\left(2y+2\right)\left(y-2\right)=19y\left(y+2\right)$

$\to y=-\dfrac{17+\sqrt{473}}{23},\:y=\dfrac{\sqrt{473}-17}{23}$

$\to |x-10|+\dfrac{|-\dfrac{17+\sqrt{473}}{23}-2|}{-\dfrac{17+\sqrt{473}}{23}}=12$

$\to x=-\dfrac{35+\sqrt{473}}{4}+10\quad \mathrm{hoặc}\quad \:x=22+\dfrac{\sqrt{473}-13}{4}$

Hoặc $|x-10|+\dfrac{|\dfrac{\sqrt{473}-17}{23}-2|}{\dfrac{\sqrt{473}-17}{23}}=12$

$\to x=-\dfrac{35-\sqrt{473}}{4}+10\quad \mathrm{hoặc}\quad \:x=22-\dfrac{13+\sqrt{473}}{4}$