Giải thích các bước giải:
Bài 2:
a.Ta có :
$\left(-\dfrac{3}{5}\right)^3-x=\dfrac{6}{50}:3$
$\to\dfrac{-27}{125}-x=\dfrac{1}{50}$
$\to x=\dfrac{-27}{125}-\dfrac{1}{25}$
$\to x=-\dfrac{32}{125}$
b.Ta có :
$|x+\dfrac{1}{7}|=\left(\dfrac{5}{6}\right)^2-\dfrac{2}{3}$
$\to |x+\dfrac{1}{7}|=\dfrac{1}{36}$
$\to x+\dfrac17=\dfrac1{36}\to x=-\dfrac{29}{252}$
Hoặc $x+\dfrac17=-\dfrac1{36}\to x=-\dfrac{43}{252}$
c.Ta có :
$|x+\dfrac{7}{24}|=\left(\dfrac{5}{6}\right)^2-\dfrac{2}{3}$
$\to |x+\dfrac{7}{24}|=\dfrac{1}{36}$
$\to x+\dfrac{7}{24}=\dfrac{1}{36}\to x=-\dfrac{19}{72}$
Hoặc $x+\dfrac{7}{24}=-\dfrac{1}{36}\to x=-\dfrac{23}{72}$
d.Ta có :
$-\dfrac45-x:\dfrac7{10}=\dfrac9{50}:3$
$\to x:\dfrac7{10}=-\dfrac45-\dfrac9{50}:3$
$\to x:\dfrac7{10}=-\dfrac45-\dfrac3{50}$
$\to x:\dfrac7{10}=-\dfrac{43}{50}$
$\to x=-\dfrac{43}{50}.\dfrac7{10}$
$\to x=-\dfrac{301}{500}$
