Ta có:
\(\frac{{a + 2020b}}{{b + 2020c}}.\frac{{a + 2020b}}{{b + 2020c}} = \frac{{\left( {a + 2020b} \right)\left( {a + 2020b} \right)}}{{\left( {b + 2020c} \right)\left( {b + 2020c} \right)}} = {\left( {\frac{{a + 2020b}}{{b + 2020c}}} \right)^2}\)
Mà \({\left( {\frac{{a + 2020b}}{{b + 2020c}}} \right)^2} = {\left( {\frac{a}{b}} \right)^2} = {\left( {\frac{b}{c}} \right)^2}\,\,\,\,\,\,\left( 1 \right)\,\,\,\, \Rightarrow \frac{a}{b} = \frac{b}{c}\)
\(\begin{array}{l} \Rightarrow ac = {b^2}\\ \Rightarrow \frac{{ac}}{{{c^2}}} = \frac{{{b^2}}}{{{c^2}}} = {\left( {\frac{b}{c}} \right)^2} = \frac{a}{c}\,\,\,\,\,\left( 2 \right)\end{array}\)
Từ \(\left( 1 \right)\) và \(\left( 2 \right)\) Suy ra: \(\frac{a}{c} = {\left( {\frac{{a + 2020b}}{{c + 2020c}}} \right)^2} = \frac{{a + 2020b}}{{b + 2020c}}.\frac{{a + 2020b}}{{b + 2020c}}\) (đpcm).
