ĐK: \(x\ne0;2\)
\(A=\frac{3}{2-x}+\frac{2}{x}=\frac{3x+4-2x}{x\left(2-x\right)}=\frac{x+4}{2x-x^2}\)
\(\Leftrightarrow A\cdot\left(2x-x^2\right)=x+4\)
\(\Leftrightarrow2Ax-x^2A=x+4\)
\(\Leftrightarrow x^2A-2Ax+x+4=0\)
\(\Leftrightarrow x^2\cdot A-x\cdot\left(2A-1\right)+4=0\)
Coi pt trên là pt bậc hai ẩn \(x\). Để pt có nghiệm thực thì:
\(\Delta=\left(2A-1\right)^2-4\cdot A\cdot4\ge0\)
\(\Leftrightarrow4A^2-4A+1-16A\ge0\)
\(\Leftrightarrow4A^2-20A+1\ge0\)
\(\Leftrightarrow4A^2-2\cdot2A\cdot5+25-24\ge0\)
\(\Leftrightarrow\left(2A-5\right)^2\ge24\)
\(\Leftrightarrow\left[{}\begin{matrix}2A-5\ge\sqrt{24}\\2A-5\le-\sqrt{24}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}A\ge\frac{2\sqrt{6}+5}{2}\\A\le\frac{-2\sqrt{6}+5}{2}\end{matrix}\right.\)
Vậy \(minA=\frac{2\sqrt{6}+5}{2}\Leftrightarrow x=2\sqrt{6}-4\)
Anh Trần Thanh Phương quá phức tạp:(
\( A=\frac{\sqrt{3}^2}{2-x}+\frac{\sqrt{2}^2}{x}\left(\text{theo đề đã sửa}\right)\ge\frac{\left(\sqrt{3}+\sqrt{2}\right)^2}{2}=\frac{5+2\sqrt{6}}{2}\) (áp dụng bđt cauchy-Schwarz)
