Giải thích các bước giải:
$\dfrac{x+3}{y-1}=\dfrac{5}{4}$
$⇔4(x+3)=5(y-1)$
$⇔4x+12=5y-1$
$⇔4x-5y+13=0$
$b)\dfrac{x+2}{0,2}=\dfrac{3,2}{x+2}$
$⇔(x+2)^2=(0,2).(3,2)$
$⇔(x+2)^2=0,64$
$⇔(x+2)^2=(±0,8)^2$
$⇔x+2=±0,8$
$⇔$ \(\left[ \begin{array}{l}x+2=0,8\\x+2=-0,8\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=-1,2\\x=-2,8\end{array} \right.\)
$\text{Vậy $x∈\{-2,8;-1,2\}$}$
Học tốt!!!
