chuyển 1 ⩽ (13 − 2Z)/Z ⩽ 1 , 5 sang 3 , 71 ⩽ Z ⩽ 4 , 33 kiểu gì

0 bình luận về “chuyển 1 ⩽ (13 − 2Z)/Z ⩽ 1 , 5 sang 3 , 71 ⩽ Z ⩽ 4 , 33 kiểu gì”

1. \[\begin{array}{l}
   1 \le \frac{{13 – 2z}}{z} \le \frac{3}{2} \Leftrightarrow \left\{ \begin{array}{l}
   \frac{{13 – 2z}}{z} – 1 \ge 0\\
   \frac{{13 – 2z}}{z} – \frac{3}{2} \le 0
   \end{array} \right.\\
   \Leftrightarrow \left\{ \begin{array}{l}
   \frac{{13 – 2z – z}}{z} \ge 0\\
   \frac{{26 – 4z – 3z}}{{2z}} \le 0
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   \frac{{13 – 3z}}{z} \ge 0\\
   \frac{{26 – 7z}}{{2z}} \le 0
   \end{array} \right.\\
   \Leftrightarrow \left\{ \begin{array}{l}
   \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   13 – 3z \ge 0\\
   z > 0
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   13 – 3z \le 0\\
   z < 0 \end{array} \right. \end{array} \right.\\ \left[ \begin{array}{l} \left\{ \begin{array}{l} 26 - 7z \ge 0\\ 2z < 0 \end{array} \right.\\ \left\{ \begin{array}{l} 26 - 7z \le 0\\ 2z > 0
   \end{array} \right.
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   z \le \frac{{13}}{2}\\
   z > 0
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   z \ge \frac{{13}}{3}\\
   z < 0 \end{array} \right. \end{array} \right.\\ \left[ \begin{array}{l} \left\{ \begin{array}{l} z \le \frac{{26}}{7}\\ z < 0 \end{array} \right.\\ \left\{ \begin{array}{l} z \ge \frac{{26}}{7}\\ z > 0
   \end{array} \right.
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   0 < z \le \frac{{13}}{3}\\ \left[ \begin{array}{l} z < 0\\ z > \frac{{26}}{7}
   \end{array} \right.
   \end{array} \right.\\
   \Leftrightarrow \frac{{26}}{7} < z \le \frac{{13}}{3} \Leftrightarrow 3,71 < z \le 4,3. \end{array}\]

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2. Đáp án:

    

   Giải thích các bước giải:

   Ca(NO3)2→Ca2++2NO−3(NH4)2SO4→2NH4++SO42−FeSO4→Fe2++SO42−HClO⇆H++ClO−NH4NO3→NH+4+NO−3Na3PO4→3Na++PO3−4Al2(SO4)3→2Al3++3SO2−4CH3COOK→CH3COO−+K+(HCOO)2Ca→2HCOO−+Ca2+HBr→H++Br−H3PO4⇆H++H2PO−4H2PO−4⇆H++HPO2−4HPO2−4⇄H+PO3−4

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