a. Điều kiện xác định: \(\left\{{}\begin{matrix}x^2-4\ne0\\6-3x\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
\(G=\left(\dfrac{x}{x^2-4}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\dfrac{6}{x+2}\)
\(=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x+2}\right]:\dfrac{6}{x+2}\)
\(=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{6}{x+2}\)
\(=\dfrac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}\)
\(=\dfrac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-1}{x-2}\)
b. Theo giả thiết, ta có:
\(\left|2x-1\right|=3\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
So với điều kiện xác định thấy \(x=-1\) thỏa mãn.
Thay \(x=-1\) vào biểu thức \(G\) ta có:
\(G=\dfrac{-1}{-1-2}=\dfrac{-1}{-3}=\dfrac{1}{3}\)
c. Ta có: \(A=\left(x^3-8\right).\dfrac{1}{x-2}+x\)
\(=\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x-2}+x\)
\(=\left(x^2+2x+4\right)+x\)
\(=x^2+3x+4\)
\(=\left(x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)
\(=\left(x+1,5\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=-1,5\)
Vậy \(MinA=\dfrac{7}{4}\Leftrightarrow x=-1,5\)
