Toán Lớp 8: 1/ |2x -3| – x = |2-x| 2/ 2|x-3|-|4x-1| = 0 3/ Tìm x,y biết : |x+2|+|2y+3|=0

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2. \(\begin{array}{l}
   1)\quad |2x -3| – x = |2-x|\qquad (1)\\
   +)\quad x >2\\
   (1)\Leftrightarrow 2x – 3 – x = x – 2\\
   \Leftrightarrow x – 3 = x – 2\\
   \Leftrightarrow -3 = -2\quad \text{(vô lí)}\\
   +)\quad \dfrac{3}{2}\leqslant x \leqslant 2\\
   (1)\Leftrightarrow 2x – 3 – x = 2 – x\\
   \Leftrightarrow 2x = 5\\
   \Leftrightarrow x = \dfrac52\quad (loại)\\
   +)\quad x < \dfrac32\\
   (1)\Leftrightarrow 3 – 2x – x = 2 – x\\
   \Leftrightarrow 1 = 2x\\
   \Leftrightarrow x = \dfrac12\quad (nhận)\\
   \text{Vậy}\ S = \left\{\dfrac12\right\}\\
   2)\quad 2|x-3| – |4x-1| = 0\\
   \Leftrightarrow 2|x-3| = |4x-1|\\
   \Leftrightarrow\left[\begin{array}{l}2(x-3) = 4x – 1\\2(x-3) = – 4x  +1\end{array}\right.\\
   \Leftrightarrow\left[\begin{array}{l}2x = -5\\6x=7\end{array}\right.\\
   \Leftrightarrow\left[\begin{array}{l}x = -\dfrac52\\x=\dfrac76\end{array}\right.\\
   \text{Vậy}\ S = \left\{-\dfrac52;\dfrac76\right\}\\
   3)\quad |x+2| + |2y + 3| = 0\qquad (3)\\
   \text{Ta có:}\\
   \begin{cases}|x+2| \geqslant 0\quad\forall x\\|2y + 3| \geqslant 0\quad \forall y\end{cases}\\
   \text{Do đó:}\\
   (3)\Leftrightarrow |x+2| = |2y + 3| = 0\\
   \Leftrightarrow \begin{cases}x  +2= 0\\2y + 3 =0\end{cases}\\
   \Leftrightarrow \begin{cases}x = -2\\y = -\dfrac32\end{cases}\\
   \text{Vậy}\ (x;y) = \left(-2;-\dfrac32\right)
   \end{array}\)

    

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