\(A=\frac{4\left(x^2-2x+1\right)+\left(x^2+2x+1\right)}{x^2-2x+1}\) \(=4+\frac{\left(x+1\right)^2}{\left(x-1\right)^2}\ge4\forall x\)
Dấu "=" \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
Vậy \(A_{min}=4\Leftrightarrow x=-1\)
\(A=\frac{5x^2-6x+5}{x^2-2x+1}\)
\(\Leftrightarrow\)\(Ax^2-5x^2-2Ax+6x+A-5=0\)
\(\Leftrightarrow\)\(\left(A-5\right)x^2-2\left(A-3\right)x+\left(A-5\right)=0\)
+) Nếu \(A=5\) thì \(x=0\)
+) Nếu \(A\ne5\) thì pt có nghiệm \(\Leftrightarrow\)\(\Delta\ge0\)
\(\Leftrightarrow\)\(\left(3-A\right)^2-\left(A-5\right)^2\ge0\)
\(\Leftrightarrow\)\(\left(3-A-A+5\right)\left(3-A+A-5\right)\ge0\)
\(\Leftrightarrow\)\(-2\left(-2A+8\right)\ge0\)
\(\Leftrightarrow\)\(A\ge4\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-1\)
Cách này thường dùng để tìm Min + Max nhé
