Giải thích các bước giải:
Câu 1:
$a)\dfrac{9}{4}+\dfrac{3}{4}:(\dfrac{2}{3}-\dfrac{5}{9})$
$=\dfrac{9}{4}+\dfrac{3}{4}:\dfrac{1}{9}$
$=\dfrac{9}{4}+\dfrac{3}{4}.9$
$=\dfrac{9}{4}+\dfrac{27}{4}$
$=9$
$b)\dfrac{2^{15}.9^4}{6^6.8^3}$
$=\dfrac{2^{15}.3^{8}}{3^6.2^6.2^{9}}$
$=\dfrac{2^{15}.3^8}{3^6.2^{15}}$
$=3^2$
$=9$
$c)\sqrt{(-5)^2}-(2020^0)^{2021}-|-\dfrac{1}{2}|$
$=5-1^{2021}-\dfrac{1}{2}$
$=5-1-\dfrac{1}{2}$
$=\dfrac{7}{2}$
Câu 2:
$a)\dfrac{2}{3}-(\dfrac{3}{4}+x)=\sqrt{\dfrac{1}{9}}$
$⇒\dfrac{2}{3}-\dfrac{3}{4}-x=\dfrac{1}{3}$
$⇒-\dfrac{1}{12}-x=\dfrac{1}{3}$
$⇒-x=\dfrac{5}{12}$
$⇒x=-\dfrac{5}{12}$
Vậy $x=-\dfrac{5}{12}$
$b)|x-\dfrac{3}{4}|-0,5=7$
$⇒|x-\dfrac{3}{4}|=\dfrac{15}{2}$
$⇒\left[ \begin{array}{l}x-\dfrac{3}{4}=\dfrac{15}{2}\\x-\dfrac{3}{4}=-\dfrac{15}{2}\end{array} \right.⇒\left[ \begin{array}{l}x=\dfrac{33}{4}\\x=-\dfrac{27}{4}\end{array} \right.$
Vậy $x∈\{-\dfrac{27}{4};\dfrac{33}{4}\}$
