4-I 2/3x+1 I = 6 (1/2.x+1).3-(1/2.x+1) :4.x=0 I 1/X – 1/3 I – 1/2 = 1/3

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1. Đáp án:

   $4-\left | \dfrac{2}{3}x+1 \right |=6\Leftrightarrow x\in \varnothing$

   $\left ( \dfrac{1}{2}x+1 \right ).3-\left ( \dfrac{1}{2}x+1 \right ):4.x=0\Leftrightarrow x=-2$

   $\left | \dfrac{1}{x}-\dfrac{1}{3} \right |-\dfrac{1}{2}=\dfrac{1}{3}\Leftrightarrow x\in \left \{ \dfrac{6}{7};-2 \right \}$

   Giải thích các bước giải:

   $4-\left | \dfrac{2}{3}x+1 \right |=6$
   $\Leftrightarrow \left | \dfrac{2}{3}x+1 \right |=-2$
   $\Leftrightarrow x\in \varnothing$ (vì $\left | \dfrac{2}{3}x+1 \right |\geq 0\forall x$)
   $\left ( \dfrac{1}{2}x+1 \right ).3-\left ( \dfrac{1}{2}x+1 \right ):4.x=0$
   $\Leftrightarrow \left ( \dfrac{1}{2}x+1 \right )\left ( 3-\dfrac{x}{4} \right )=0$
   $\Leftrightarrow \dfrac{1}{2}x+1=0$
   $\Leftrightarrow \dfrac{1}{2}x=-1$
   $\Leftrightarrow x=-2$
   $\left | \dfrac{1}{x}-\dfrac{1}{3} \right |-\dfrac{1}{2}=\dfrac{1}{3}$
   $\Leftrightarrow \left | \dfrac{1}{x}-\dfrac{1}{3} \right |=\dfrac{5}{6}$
   $\Leftrightarrow \left[ \begin{array}{l}\dfrac{1}{x}-\dfrac{1}{3}=\dfrac{5}{6}\\\dfrac{1}{x}-\dfrac{1}{3}=-\dfrac{5}{6}\end{array} \right.$
   $\Leftrightarrow \left[ \begin{array}{l}\dfrac{1}{x}=\dfrac{7}{6}\\\dfrac{1}{x}=-\dfrac{1}{2}\end{array} \right.$
   $\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{6}{7}\\x=-2\end{array} \right.$

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