$a$) $(x-5)^2= (1-3x)^2$
$⇒$ \(\left[ \begin{array}{l}x-5=1-3x\\x-5=-(1-3x)\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x+3x=1+5\\x-5=-1+3x)\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}4x=6\\x-3x=-1+5)\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=-2)\end{array} \right.\)
Vậy $x$ $∈$ `{\frac{3}{2};-2}`
$b$) $x^2+x=0$
$⇔ x.(x+1)=0$
$⇒$ \(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
Vậy $x$ $∈$ {$0;-1$}
$c$) $3^4 x 3^n = 3^7$
$⇔ 3^{4+n} = 3^7$
$⇔ 4+n=7$
$⇔ n=3$
Vậy $n=3$.
