` 8/23 . 46/24 =1/3 .x`

`=>8/23 . 23/12 =1/3 . x`

`=> 1/3 . x=2/3`

`=>x=2/3 : 1/3`

`=>x=2/3 . 3`

`=> x= 6/3`

`=>x=2`

`----`

`1/5 : x= 1/5-1/7`

`=>1/5 : x=  7/35 - 5/35`

`=> 1/5 :x= 2/35`

`=>x= 1/5 : 2/35`

`=>x=1/5 . 35/2`

`=>x=7/2`

`----`

`4/9 - (x-1/2)^2 =1/3`

`=> (x-1/2)^2 =4/9-1/3`

`=> (x-1/2)^2 =4/9- 3/9`

`=> (x-1/2)^2 =1/9`

`=> (x-1/2)^2 = (+- 1/3)^2`

`@ TH1`

`x-1/2=1/3`

`=>x=1/3+1/2`

`=>x= 2/6 + 3/6`

``=>x= 5/6`

`@ TH2`

`x-1/2=-1/3`

`=>x=-1/3 +1/2`

`=>x= -2/6 + 3/6`

`=>x=1/6`

`----`

`3,2 . x-(4/5+2/3) : 3 2/3 = 7/10`

`=> 3,2 . x-22/15 : 11/3 = 7/10`

`=>  3,2 . x-22/15 = 7/10 . 11/3`

`=>  3,2 . x-22/15 =77/30`

`=> 3,2 .x= 77/30 + 22/15`

`=> 3,2 .x=121/30`

`=>x= 121/30. 5/16`

`=>x= 121/96`

\(3,2x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):3\dfrac{2}{3}=\dfrac{7}{20}\\ \Rightarrow\dfrac{16}{5}x-\dfrac{22}{15}:\dfrac{11}{3}=\dfrac{7}{20}\\ \Rightarrow\dfrac{16}{5}x-\dfrac{2}{5}=\dfrac{7}{20}\\ \Rightarrow\dfrac{16}{5}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{15}{64}\)

\(\left(4\dfrac{1}{2}-2x\right).1\dfrac{4}{61}=6\dfrac{1}{2}\\ \Rightarrow\left(\dfrac{9}{2}-2x\right).\dfrac{65}{61}=\dfrac{13}{2}\\ \Rightarrow\dfrac{9}{2}-2x=\dfrac{61}{10}\\ \Rightarrow2x= -\dfrac{8}{5}\\ \Rightarrow x=-\dfrac{4}{5}\)

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

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

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

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

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

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

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

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

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

\(\left|x-\frac{1}{3}\right|=\frac{10}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{-10}{5}\\x-\frac{1}{3}=\frac{10}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{3}\\x=\frac{7}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-5}{3};\frac{7}{3}\right\}\)

Chúc bạn học tốt nhé !!!

∣ ∣ ∣ x − 1 3 ∣ ∣ ∣ + 4 5 = ∣ ∣ ∣ − 3.2 + 2 5 ∣ ∣ ∣ ⇒ ∣ ∣ ∣ x − 1 3 ∣ ∣ ∣ + 4 5 = ∣ ∣ ∣ − 28 5 ∣ ∣ ∣ ⇒ ∣ ∣ ∣ x − 1 3 ∣ ∣ ∣ + 4 5 = 28 5 ⇒ ∣ ∣ ∣ x − 1 3 ∣ ∣ ∣ = 28 5 − 4 5 = 24 5 ⇒ x − 1 3 = ( ± 24 5 ) ⇒ ⎡ ⎢ ⎢ ⎣ x − 1 3 = 24 5 x − 1 3 = − 24 5 ⇒ ⎡ ⎢ ⎢ ⎣ x = 24 5 + 1 3 x = − 24 5 + 1 3 ⇒ ⎡ ⎢ ⎢ ⎣ x = 77 15 x = − 67 15

\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\left(-3,2\right)+\dfrac{2}{5}\right|\\ \Rightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\dfrac{-14}{5}\right|\\ \Rightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\dfrac{14}{5}\\ \Rightarrow\left|x-\dfrac{1}{3}\right|=2\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=-2\\x-\dfrac{1}{3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\\x=\dfrac{7}{3}\end{matrix}\right.\)

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

\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left[-\frac{3}{2}+\frac{2}{5}\right]\)

\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{4}{5}=-\frac{11}{10}\)

\(\Rightarrow\left|x-\frac{1}{3}\right|=-\frac{11}{10}-\frac{4}{5}\)

\(\Rightarrow\left|x-\frac{1}{3}\right|=-\frac{19}{10}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{19}{10}\\x-\frac{1}{3}=-\frac{19}{10}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{67}{30}\\x=-\frac{47}{30}\end{cases}}\)

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