3(x-1/2)^2-1/3=0 giups minhf aj

2 bình luận về “3(x-1/2)^2-1/3=0 giups minhf aj”

1. 3(x-1/2)^2 -1/3=0

   =>3(x-1/2)^2 =0+1/3

   =>3(x-1/2)^2 =1/3

   =>(x-1/2)^2 =1/3:3

   =>(x-1/2)^2 =1/9

   =>(x-1/2)^2 =(+-1/3)^2

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

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

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

   Vậy x\in{5/6;1/6}

    

   Trả lời
2. 3(x-1/2)^2-1/3=0

   => 3(x-1/2)^2=1/3

   => (x-1/2)^2=1/3:3

   => (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=5/6

   TH2: x-1/2=-1/3

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

   => x=1/6

   Vậy x\in{5/6;1/6}

   \color{orange}{\text {#moc}

    

   Trả lời