`a, 3,2 \div x=0,8`

`x=3,2 \div 0,8`

`x=4`

`b, 5 \times x = 37,5`

`x=37,5 \div 5`

`x=7,5`

`@` `\text {_vdn_}`

\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left(-3,2\right)+\dfrac{2}{5}\)

\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=-\dfrac{14}{5}\)

\(\left|x-\dfrac{1}{3}\right|=-\dfrac{14}{5}-\dfrac{4}{5}=-\dfrac{18}{5}\)

Vì \(\left|x-\dfrac{1}{3}\right|\ge0\) ∀x

⇒Phương trình vô nghiệm

|x-\(\dfrac{1}{3}\)|+\(\dfrac{4}{5}\)=|(\(\dfrac{-16}{5}\))+\(\dfrac{2}{5}\)|

⇒|x-\(\dfrac{1}{3}\)|+\(\dfrac{4}{5}\)=|\(\dfrac{-14}{5}\)|

⇒|x-\(\dfrac{1}{3}\)|+\(\dfrac{4}{5}\)=\(\dfrac{14}{5}\)

⇒|x-\(\dfrac{1}{3}\)|=\(\dfrac{14}{5}\)-\(\dfrac{4}{5}\)

⇒|x-\(\dfrac{1}{3}\)|=2

⇒x-\(\dfrac{1}{3}\)=2⇒x=\(\dfrac{7}{3}\)

hoặc

⇒x-\(\dfrac{1}{3}\)=-2⇒x=\(\dfrac{-5}{3}\)

Vậy x=\(\dfrac{7}{3}\) hoặc x=\(\dfrac{-5}{3}\)

\(\dfrac{-5}{3}\)

=> |x-1/3| +4/5 = 0,7

=> |x-1/3|  = -3,8

mà |x-1/3|  \(\ge\)0

=> ko tồn tại x

mà[x-1/3] > 0 nha

\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|-3,2+\dfrac{2}{5}\right|\)

=>\(\left|x-\dfrac{1}{3}\right|+0,8=\left|-3,2+0,4\right|=2,8\)

=>\(\left|x-\dfrac{1}{3}\right|=2,8-0,8=2\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

3,2 x 4 + 2 x 32 + 3,2 x 4

= 3,2 x (4+2+4)

=3,2 x 10 

=32

   3,2 : 0,25 + 2 x 3,2 + 3,2 x 100 : 25

= 3,2 x 4 + 3,2 x 2 + 3,2 x 4

= 3,2 x ( 4 + 2 + 4)

= 3,2 x 10

= 32

$3,2\times x+3,26=17,66$

$3,2\times x=17,66-3,26$

$3,2\times x=14,4$

$x=14,4:3,2$

$x=4,5$

3,2 x X + 3,26 = 17,66

3,2 x X = 17,66 - 3,26

3,2 x X = 14,4

X = 14,4 : 3,2

X = 4,5