Toán Lớp 6: 2 . |-3 + 1| – |1/2 – x| = 3
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Toán Lớp 6: 2 . |-3 + 1| – |1/2 – x| = 3
Tìm x
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$\text{nội dung}$
2. |-3 + 1| – |\frac{1}{2} – x| = 3
⇒ 2. |-2| – |\frac{1}{2} – x| = 3
⇒ 2. 2 – |\frac{1}{2} – x| = 3
⇒ 4 – |\frac{1}{2} – x| = 3
⇒ |\frac{1}{2} – x| = 1
TH1: \frac{1}{2} – x = 1 (x \le \frac{1}{2})
⇒ x = \frac{1}{2} – 1
⇒ x = \frac{-1}{2} (TMĐK)
TH2: x – \frac{1}{2} = 1 (x > \frac{1}{2})
⇒ x = 1 + \frac{1}{2}
⇒ x = \frac{3}{2} (TMĐK)
Vậy x = \frac{-1}{2} và x = \frac{3}{2}
2. |-3+1| – |1/2 – x| = 3
=> 2. |-2| – |1/2 – x| = 3
=> 2.2 – |1/2 – x| = 3
=> 4 – |1/2 – x| = 3
=> |1/2 – x| = 1
=> \(\left[ \begin{array}{l}\dfrac{1}{2}-x=1\\\dfrac{1}{2}-x=-1\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=\dfrac{1}{2}-1\\x=\dfrac{1}{2}-(-1)\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=\dfrac{-1}{2}\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy x in {-1/2; 3/2}
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