a) $(5x-1)^2=\dfrac{36}{49}$
$↔ \left[ \begin{array}{l}5x-1=-\dfrac{6}{7}\\5x-1=\dfrac{6}{7}\end{array} \right.$
$↔ \left[ \begin{array}{l}x=\dfrac{1}{35}\\x=\dfrac{13}{35}\end{array} \right.$
b) $\Bigg(x-\dfrac{1}{9}\Bigg)^2=\Bigg(\dfrac{2}{3}\Bigg)^6$
$↔ \Bigg(x-\dfrac{1}{9}\Bigg)^2=\Bigg(\dfrac{8}{27}\Bigg)^2$
$↔ \left[ \begin{array}{l}x-\dfrac{1}{9}=\dfrac{8}{27}\\x-\dfrac{1}{9}=-\dfrac{8}{27}\end{array} \right.$
$↔ \left[ \begin{array}{l}x=\dfrac{11}{27}\\x=-\dfrac{5}{27}\end{array} \right.$
c) $(8x-1)^{2x+1}=5^{2x+1}$
$↔ 8x-1=5$
$↔ 8x=6$
$↔ x=\dfrac{3}{4}$
d) $(x-3,5)^2+\Bigg(y-\dfrac{1}{10}\Bigg)^4≤0$
Vì $(x-3,5)^2≥0$, $\Bigg(y-\dfrac{1}{10}\Bigg)^4≥0$ nên:
$(x-3,5)^2+\Bigg(y-\dfrac{1}{10}\Bigg)^4≤0$
$↔ \left[ \begin{array}{l}x-3,5=0\\y-\dfrac{1}{10}=0\end{array} \right.$
$↔ \left[ \begin{array}{l}x=3,5\\y=\dfrac{1}{10}\end{array} \right.$
