Đáp án:
9) \(T = \dfrac{{x + 20}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
8)\left( {2a - 5} \right)\left( {2a + 5} \right)x = 5a + 5\\
\to x = \dfrac{{5a + 5}}{{\left( {2a - 5} \right)\left( {2a + 5} \right)}}\\
5\left( {a - 1} \right)\left( {a + 1} \right)y = 3a\left( {2a + 5} \right)\\
\to y = \dfrac{{3a\left( {2a + 5} \right)}}{{5\left( {a - 1} \right)\left( {a + 1} \right)}}\\
A = xy = \dfrac{{5\left( {a + 1} \right)}}{{\left( {2a - 5} \right)\left( {2a + 5} \right)}}.\dfrac{{3a\left( {2a + 5} \right)}}{{5\left( {a - 1} \right)\left( {a + 1} \right)}}\\
= \dfrac{{3a}}{{\left( {2a - 5} \right)\left( {a - 1} \right)}}\\
9)\dfrac{1}{x}.\dfrac{x}{{x + 2}}.\dfrac{{x + 2}}{{x + 4}}.\dfrac{{x + 4}}{{x + 6}}....\dfrac{{x + 16}}{{x + 18}}.\dfrac{{x + 18}}{{x + 20}}.T = \dfrac{1}{2}\\
\to \dfrac{1}{{x + 20}}.T = \dfrac{1}{2}\\
\to 2T = x + 20\\
\to T = \dfrac{{x + 20}}{2}
\end{array}\)
