Tìm x bt `a,(5x-1)(2x-1/3)=0` `b,(2x-8)(3/4x+1/8)=0

2 bình luận về “Tìm x bt `a,(5x-1)(2x-1/3)=0` `b,(2x-8)(3/4x+1/8)=0”

1. a,(5x-1)(2x-1/3)=0

   => \(\left[ \begin{array}{l}5x-1=0\\2x-1/3=0\end{array} \right.\)

   => \(\left[ \begin{array}{l}5x=0+1\\x=2x=0+1/3\end{array} \right.\) 

   => \(\left[ \begin{array}{l}5x=1\\2x=1/3\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x=1:5\\x=1/3:2\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x=1/5\\x=1/6\end{array} \right.\) 

   Vậy x in {1/5;1/6}

   b,(2x-8)(3/4x+1/8)=0

   => \(\left[ \begin{array}{l}2x-8=0\\3/4.x+1/8=0\end{array} \right.\) 

   => \(\left[ \begin{array}{l}2x=8\\3/4x=-1/8\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x=8:2\\x=-1/8:3/4\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x=4\\x=-1/6\end{array} \right.\) 

   Vậy x in {-1/6;4}

   Trả lời
2. Giải đáp:

   a, (5x-1)(2x-1/3)=0

   =>\(\left[ \begin{array}{l}5x-1=0\\2x-\dfrac{1}{3}=0\end{array} \right.\) 

   =>\(\left[ \begin{array}{l}5x=1\\2x=\dfrac{1}{3}\end{array} \right.\) 

   =>\(\left[ \begin{array}{l}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{array} \right.\) 

   Vậy x\in{1/5;1/6}

   b, (2x-8)(3/4x+1/8)=0

   =>\(\left[ \begin{array}{l}2x-8=0\\\dfrac{3}{4}x+\dfrac{1}{8}=0\end{array} \right.\) 

   =>\(\left[ \begin{array}{l}2x=8\\\dfrac{3}{4}x=-\dfrac{1}{8}\end{array} \right.\) 

   =>\(\left[ \begin{array}{l}x=4\\x=-\dfrac{1}{6}\end{array} \right.\) 

   Vậy x\in{4;-1/6}

    

   Trả lời