1 tìm x a, ( x -1 ) ( x + 1/ 2 ) = 0 b , ( 2 x – 1 ) ( x + 2/5 ) = 0 2

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1. ${a,(x-1)(x+}$ $\dfrac{1}{2})=0$ 

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

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

   Vậy ${x}$ ∈ {${1;}$ $-\dfrac{1}{2}$}

   ${b,(2x-1)(x+}$ $\dfrac{2}{5})=0$

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

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

   Vậy ${x}$ ∈ {${}$ $\dfrac{1}{2};-$ $\dfrac{2}{5}$}

      ~Xin hay nhất~

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2. Đáp án:

   a) $x\in \left \{ 1;\dfrac{1}{2} \right \}$

   b) $x\in \left \{ \dfrac{1}{2};-\dfrac{2}{5} \right \}$

   Giải thích các bước giải:

   a) $(x-1)\left ( x+\dfrac{1}{2} \right )=0$
   $\Leftrightarrow \left[ \begin{array}{l}x-1=0\\x+\dfrac{1}{2}=0\end{array} \right.$
   $\Leftrightarrow \left[ \begin{array}{l}x=1\\x=-\dfrac{1}{2}\end{array} \right.$
   Vậy $x\in \left \{ 1;\dfrac{1}{2} \right \}$
   b) $(2x-1)\left ( x+\dfrac{2}{5} \right )=0$
   $\Leftrightarrow \left[ \begin{array}{l}2x-1=0\\x+\dfrac{2}{5}=0\end{array} \right.$
   $\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{2}{5}\end{array} \right.$
   Vậy $x\in \left \{ \dfrac{1}{2};-\dfrac{2}{5} \right \}$

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