Ta có: \(B=5-\dfrac{\left(x+19\right)^2-325}{3\left(x+1\right)^2+6}=\dfrac{14x^2-8x+9}{3x^2+6x+9}\) (nhân vào rồi rút gọn)
\(\Rightarrow3Bx^2+6Bx+9B=14x^2-8x+9\)
\(\Rightarrow3Bx^2+6Bx+9B-14x^2+8x-9=0\)
\(\Leftrightarrow\left(3B-14\right)x^2+\left(6B+8\right)x+\left(9B-9\right)=0\) (1)
Quan niệm (1) là phương trình bậc 2 ẩn x
\(\Rightarrow\Delta=\left(6B+8\right)^2-4\left(3B-14\right)\left(9B-9\right)\ge0\)
\(\Leftrightarrow\Delta=36B^2+96B+64-108B^2+612B-504\ge0\)
\(\Leftrightarrow-72B^2+708B-440\ge0\)
\(\Leftrightarrow72B^2-708B+440\le0\)
\(\Leftrightarrow72\left(B-\dfrac{55}{6}\right)\left(B-\dfrac{2}{3}\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}B-\dfrac{55}{6}\ge0\\B-\dfrac{2}{3}\le0\end{matrix}\right.\\\left\{{}\begin{matrix}B-\dfrac{55}{6}\le0\\B-\dfrac{2}{3}\ge0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}B\ge\dfrac{55}{6}\\B\le\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}B\le\dfrac{55}{6}\\B\ge\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Chọn \(\dfrac{2}{3}\le B\le\dfrac{55}{6}\) Vậy GTNN của B là: \(\dfrac{2}{3}\)
