ĐKXĐ : \(x\ne0\)

Ta có : \(A=\frac{x^2-2x+2014}{x^2}\)

\(\Rightarrow2014A=\frac{2014x^2-2\cdot x\cdot2014+2014^2}{x^2}\)

\(=2014-2\cdot\frac{2014}{x}+\left(\frac{2014}{x}\right)^2\)

\(=2013+\left(\frac{2014}{x}-1\right)^2\)

Ta thấy : \(\left(\frac{2014}{x}-1\right)^2\ge0\forall x\)

\(\Rightarrow2013+\left(\frac{2014}{x}-1\right)^2\ge2013\forall x\)

hay : \(2014A\ge2013\)

\(\Leftrightarrow A\ge\frac{2013}{2014}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(\frac{2014}{x}-1\right)^2=0\)

\(\Leftrightarrow x=2014\) ( thỏa mãn ĐKXĐ )

Vậy  \(min\) \(A=\frac{2013}{2014}\) tại \(x=2014\)

\(B=a^3+b^3+ab=\left(a+b\right)\left(a^2-ab+b^2\right)+ab\)

Với \(a+b=1\)ta có: \(B=a^2-ab+b^2+ab=a^2+b^2\)\

Từ \(a+b=1\)\(\Rightarrow b=1-a\)

\(\Rightarrow B=a^2+\left(1-a\right)^2=a^2+1-2a+a^2=2a^2-2a+1\)

\(=2\left(a^2-a+\frac{1}{4}\right)+\frac{1}{2}=2\left(a^2-2.\frac{1}{2}a+\frac{1}{4}\right)+\frac{1}{2}\)

\(=2\left(a-\frac{1}{2}\right)^2+\frac{1}{2}\)

Vì \(\left(a-\frac{1}{2}\right)^2\ge0\forall a\)\(\Rightarrow2\left(a-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall a\)

hay \(B\ge\frac{1}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow a-\frac{1}{2}=0\)\(\Leftrightarrow a=\frac{1}{2}\)

\(\Rightarrow b=1-\frac{1}{2}=\frac{1}{2}\)

Vậy \(minB=\frac{1}{2}\)\(\Leftrightarrow a=b=\frac{1}{2}\)