a) ( 5x – $1^{2}$ )= $\frac{36}{49}$ b) ( x – $\frac{1}{9}$$)^{2}$ ) =( $\frac{2}{3}$ $)^{6}$ c) ( 8x – $1)^{2x+1}$ = $5^{2x+1}$ d) (x-3,5 $)^{2

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:

$(x-3,5)^2+\Bigg(y-\dfrac{1}{10}\Bigg)^4≤0$

$↔ \left[ \begin{array}{l}x-3,5=0\\y-\dfrac{1}{10}=0\end{array} \right.$

$↔ \left[ \begin{array}{l}x=3,5\\y=\dfrac{1}{10}\end{array} \right.$