Ta có: \(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}=42^5:\left(2^3\cdot21^6\right)\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5\cdot21^5}{2^3\cdot21^5\cdot21}\)
\(\Leftrightarrow\left(x-2\right)\cdot\frac{4}{21}=\frac{4}{21}\)
\(\Leftrightarrow x-2=1\)
hay x=3
Vậy: x=3