Bài 1: a) /2x – 5/=4 b)1/3 – /5/4 – 2x/=1/4 c)3/4 – /2x + 1/=7/8 d)/x – 1,5/= 2 e)/x + 1/ /y – 3/=0

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1. a, $/2x – 5/=4 $ 

     \(⇔\left[ \begin{array}{l}2x – 5=4 \\2x-5=-4\end{array} \right.\) 

   \(⇔\left[ \begin{array}{l}2x=9\\2x=1\end{array} \right.\) 

   \(⇔\left[ \begin{array}{l}x=\dfrac{9}{2} \\x=\dfrac{1}{2} \end{array} \right.\)

    Vậy $x=\dfrac{9}{2}$ $hoặc$ $x=\dfrac{1}{2}$ 

   b,  $\dfrac{1}{3}-$  $/\dfrac{5}{4}-2x/=$ $\dfrac{1}{4}$ 

       $⇔/\dfrac{5}{4}-2x/=$ $\dfrac{1}{12}$ 

       \(⇔\left[ \begin{array}{l}\dfrac{5}{4}-2x=\dfrac{1}{12} \\\dfrac{5}{4} -2x=\dfrac{-1}{12} \end{array} \right.\)  

       \(⇔\left[ \begin{array}{l} x=\dfrac{7}{12} \\x=\dfrac{2}{3}\end{array} \right.\)

   Vậy $x=\dfrac{7}{12}$ hoặc $x=\dfrac{2}{3}$ 

   c, $\dfrac{3}{4}-/2x+1/=$ $\dfrac{7}{8}$ 

       $⇔/2x+1/=\dfrac{-1}{8}$ 

   Vì $/2x+1/$ $\geq0$ 

   ⇒ Phương trình vô nghiệm.

   d, $/x – 1,5/= 2$ 

     \(⇔\left[ \begin{array}{l}x= \dfrac{7}{2} \\x=\dfrac{-1}{2} \end{array} \right.\) 

   Vậy $x=\dfrac{7}{2}$ hoặc $x=\dfrac{-1}{2}$ 

   e, $⇒/x+1/=0$ và $y-3=0$ 

   $⇒x=-1$ và $y=3$ 

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