Ta có: \(A=4x+\frac{25}{x-1}=4\left(x-1\right)+\frac{25}{x-1}+4\)
Do x > 1 => x - 1 > 0
Áp dụng bđt cosi cho 2 số dương 4(x - 1) và 25/(x - 1)
Ta có: \(4\left(x-1\right)+\frac{25}{x-1}\ge2\sqrt{4\left(x-1\right)\cdot\frac{25}{x-1}}=2.10=20\)
=> \(4\left(x-1\right)+\frac{25}{x-1}+4\ge20+4=24\)
Hay \(A\ge24\)
Dấu "=" xảy ra <=> \(4\left(x-1\right)=\frac{25}{x-1}\)
<=> \(\left(x-1\right)^2=\frac{25}{4}\) <=> \(\orbr{\begin{cases}x-1=\frac{5}{2}\\x-1=-\frac{5}{2}\end{cases}}\) <=> \(\orbr{\begin{cases}x=\frac{7}{2}\left(tm\right)\\x=-\frac{3}{2}\left(ktm\right)\end{cases}}\)
Vậy MinA = 24 khi x = 7/2