Giải thích các bước giải:
\(\begin{array}{l}
a)\,{\left( {x - 7} \right)^{x + 1}} = {\left( {x - 7} \right)^{x + 11}}\\
TH1:\,x - 7 = 0 \Leftrightarrow x = 7\\
TH2:\,x - 7 = 1 \Leftrightarrow = 8\\
b)\,Vi\,\left| {3x - 4} \right| \ge 0;\,\left| {3y + 5} \right| \ge 0\\
\Rightarrow \left| {3x - 4} \right| + \left| {3y + 5} \right| = 0\\
\Leftrightarrow 3x - 4 = 0\,va\,3x + 5 = 0\\
\Leftrightarrow x = \dfrac{4}{3}\,va\,y = - \dfrac{5}{3}\\
c)\,Vi\,\,\left| {x + \dfrac{3}{4}} \right| \ge 0;\,\left| {y - \dfrac{1}{5}} \right| \ge 0;\left| {x + y + z} \right| \ge 0\\
\Rightarrow \left| {x + \dfrac{3}{4}} \right| + \left| {y - \dfrac{1}{5}} \right| + \left| {x + y + z} \right| = 0\\
\Leftrightarrow x + \dfrac{3}{4} = 0\,va\,y - \dfrac{1}{5} = 0\,va\,\,x + y + z = 0\\
\Leftrightarrow x = - \dfrac{3}{4}\,va\,y = \dfrac{1}{5}\,va\, - \dfrac{3}{4} + \dfrac{1}{5} + z = 0 \Rightarrow z = \dfrac{{11}}{{20}}\\
d)\,\left| {x + \dfrac{9}{2}} \right| + \left| {y + \dfrac{4}{3}} \right| + \left| {z + \dfrac{7}{2}} \right| \le 0\\
\Rightarrow \left| {x + \dfrac{9}{2}} \right| + \left| {y + \dfrac{4}{3}} \right| + \left| {z + \dfrac{7}{2}} \right| = 0\\
\Rightarrow x = - \dfrac{9}{2};\,y = - \dfrac{4}{3};z = - \dfrac{7}{2}
\end{array}\)
