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}\)
