1/4 + 1/3 : ║2x – 1 ║ = 11/12 ║là trị tuyệt đối

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1. Đáp án:

   `1/4 + 1/3 : |2x – 1| = 11/12`

   `1/3 : |2x – 1| = 11/12 – 1/4`

   `1/3 : |2x – 1| = 2/3`

   `|2x – 1| = 1/3 : 2/3`

   `|2x – 1| = 1/2`

   `⇒` \(\left[ \begin{array}{l}2x – 1 =\frac{1}{2} \\2x – 1 =-\frac{1}{2}\end{array} \right.\)

   `⇒` \(\left[ \begin{array}{l}2x =\frac{3}{2} \\2x = \frac{1}{2}\end{array} \right.\) 

   `⇒` \(\left[ \begin{array}{l}x =\frac{3}{4} \\x = \frac{1}{4}\end{array} \right.\) 

   Vậy `x ∈ {3/4 ; 1/4}`

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2. Đáp án:

    

   Giải thích các bước giải:

   $\frac{1}{4}$ + $\frac{1}{3}$ : |2x-1| = $\frac{11}{12}$ 

   ⇒ $\frac{1}{3}$ : |2x-1| = $\frac{11}{12}$ – $\frac{1}{4}$ 

   ⇒ $\frac{1}{3}$ : |2x-1| = $\frac{11}{12}$ – $\frac{3}{12}$

   ⇒ $\frac{1}{3}$ : |2x-1| = $\frac{11-3}{12}$

   ⇒ $\frac{1}{3}$ : |2x-1| = $\frac{2}{3}$ 

   ⇒ |2x-1| = $\frac{1}{3}$ : $\frac{2}{3}$

   ⇒ |2x-1| = $\frac{1}{3}$ . $\frac{3}{2}$

   ⇒ |2x-1| = $\frac{1}{3} 

   ⇒ \(\left[ \begin{array}{l}2x – 1 = \frac{1}{3}\\2x = \frac{1}{3}\end{array} \right.\)

   ⇒ \(\left[ \begin{array}{l}2x – 1= \frac{1}{3}\\2x = -\frac{1}{3}+1\end{array} \right.\)

    

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