Question

Biết số thực a khác 0 thỏa mãn \(9\left(a+\dfrac{1}{36a}\right)^2-6\left(a+\dfrac{1}{36a}\right)+1=0\) . Khi đó \(\dfrac{1}{a}=\)

Answer 1

Đặt \(a+\dfrac{1}{36a}=x\)

pt đã cho trở thành 9x² - 6x + 1 = 0

\(\Leftrightarrow9\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow9\left(x-\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{3}=0\)

\(\Leftrightarrow x=\dfrac{1}{3}=a+\dfrac{1}{36a}=\dfrac{36a^2+1}{36a}\)

\(\Leftrightarrow12a=36a^2+1\)

\(\Leftrightarrow36a^2-12a+1=0\)

\(\Leftrightarrow\left(6a-1\right)^2=0\)

\(\Leftrightarrow6a-1=0\)

\(\Leftrightarrow a=\dfrac{1}{6}\) \(\Rightarrow a=6\)

Answer 2

6