Question

Answer 1

\(\dfrac{1}{30^2}=\dfrac{1}{\left(\dfrac{5}{6}AC\right)^2}+\dfrac{1}{AC^2}\)

\(\Leftrightarrow\dfrac{1}{900}=\dfrac{1}{\dfrac{25}{36}AC^2}+\dfrac{1}{AC^2}\left(1\right)\)

Đặt \(AC=x\left(x>0\right)\)

\(\Rightarrow AC^2=x^2\)

\(\left(1\right)\Leftrightarrow\dfrac{1}{900}=\dfrac{1}{\dfrac{25}{36}x^2}+\dfrac{1}{x^2}\)

\(\Leftrightarrow\dfrac{1}{900}=\dfrac{36}{25x^2}+\dfrac{1}{x^2}\)

\(\Leftrightarrow\dfrac{25x^4}{22500x^4}=\dfrac{32400x^2}{22500x^4}+\dfrac{22500x^2}{22500x^4}\)

\(\Rightarrow25x^4=32400x^2+22500x^2\)

\(\Leftrightarrow25x^4=54900x^2\) (chia cả hai vế cho \(25x^2\))

\(\Leftrightarrow x^2=2196\)

\(\Rightarrow AC^2=2196\)

\(\Rightarrow AC=6\sqrt{61}\)

Vậy: \(AC=6\sqrt{61}\)