B1 GPT a. x+5/x-1>=0 b.x-15/73+ x-13/71<= x-11/69+ x-9/67 B2 cho 1/x+1/y+1/z=0 tinh gia tri bieu thuc A= yz/x^2 + zx/y^2 +xy/z^2 Mn giup e vs ^^

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1. Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   1,\\
   a,\\
   \dfrac{{x + 5}}{{x – 1}} \ge 0\,\,\,\,\,\,\,\left( {x \ne 1} \right) \Leftrightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x + 5 \ge 0\\
   x – 1 > 0
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x + 5 \le 0\\
   x – 1 < 0
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x \ge  – 5\\
   x > 1
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x \le  – 5\\
   x < 1
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x > 1\\
   x \le  – 5
   \end{array} \right.\\
   b,\\
   \dfrac{{x – 15}}{{73}} + \dfrac{{x – 13}}{{71}} \le \dfrac{{x – 11}}{{69}} + \dfrac{{x – 9}}{{67}}\\
    \Leftrightarrow \left( {\dfrac{{x – 15}}{{73}} + 1} \right) + \left( {\dfrac{{x – 13}}{{71}} + 1} \right) \le \left( {\dfrac{{x – 11}}{{69}} + 1} \right) + \left( {\dfrac{{x – 9}}{{67}} + 1} \right)\\
    \Leftrightarrow \dfrac{{\left( {x – 15} \right) + 73}}{{73}} + \dfrac{{\left( {x – 13} \right) + 71}}{{71}} \le \dfrac{{\left( {x – 11} \right) + 69}}{{69}} + \dfrac{{\left( {x – 9} \right) + 67}}{{67}}\\
    \Leftrightarrow \dfrac{{x + 58}}{{73}} + \dfrac{{x + 58}}{{71}} \le \dfrac{{x + 58}}{{69}} + \dfrac{{x + 58}}{{67}}\\
    \Leftrightarrow \left( {x + 58} \right)\left( {\dfrac{1}{{73}} + \dfrac{1}{{71}} – \dfrac{1}{{69}} – \dfrac{1}{{67}}} \right) \le 0\\
   \dfrac{1}{{73}} + \dfrac{1}{{71}} < \dfrac{1}{{69}} + \dfrac{1}{{67}} \Rightarrow \dfrac{1}{{73}} + \dfrac{1}{{71}} – \dfrac{1}{{69}} – \dfrac{1}{{67}} < 0\\
    \Rightarrow x + 58 \ge 0\\
    \Leftrightarrow x \ge  – 58\\
   2,\\
   {x^3} + {y^3} + {z^3}\\
    = \left( {{x^3} + {y^3} + 3x{y^2} + 3{x^2}y} \right) + {z^3} – \left( {3x{y^2} + 3{x^2}y} \right)\\
    = {\left( {x + y} \right)^3} + {z^3} – 3xy\left( {x + y} \right)\\
    = \left[ {\left( {x + y} \right) + z} \right].\left[ {{{\left( {x + y} \right)}^2} – \left( {x + y} \right).z + {z^2}} \right] – 3xy\left( {x + y + z} \right) + 3xyz\\
    = \left( {x + y + z} \right).\left( {{x^2} + {y^2} + {z^2} + 2xy – xz – yz} \right) – 3xy\left( {x + y + z} \right) + 3xyz\\
    = \left( {x + y + z} \right).\left( {{x^2} + {y^2} + {z^2} + 2xy – xz – yz – 3xy} \right) + 3xyz\\
    = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} – xy – yz – zx} \right) + 3xyz\\
   \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0 \Leftrightarrow \dfrac{{xy + yz + zx}}{{xyz}} = 0 \Leftrightarrow xy + yz + zx = 0\\
   A = \dfrac{{yz}}{{{x^2}}} + \dfrac{{zx}}{{{y^2}}} + \dfrac{{xy}}{{{z^2}}}\\
    = \dfrac{{{y^3}{z^3} + {z^3}{x^3} + {z^3}{y^3}}}{{{x^2}{y^2}{z^2}}}\\
    = \dfrac{{\left( {xy + yz + zx} \right)\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2} – xyyz – yzzx – zxxy} \right) + 3xy.yz.zx}}{{{x^2}{y^2}{z^2}}}\\
    = \dfrac{{0.\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2} – xyyz – yzzx – zxxy} \right) + 3{x^2}{y^2}{z^2}}}{{{x^2}{y^2}{z^2}}}\\
    = 3
   \end{array}\)

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