\(x^4-2x^3+3x^2-4x+2015=\left(x^2-x\right)^2+2\left(x-1\right)^2+2013\)

Mà \(\left(x^2-x\right)^2\ge0\forall x\); \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow Min=2013\)

Dấu "=" xảy ra \(\Leftrightarrow x=1\)

Cách này cũng khá giống của bạn Nguyễn Văn Hạ nhưng mình nghĩ dễ bến đối hơn chỗ \(x^4-2x^3+x^2\rightarrow x^2\left(x-1\right)^2\)

\(A=\left(x^4-2x^3+x^2\right)+\left(2x^2-4x+2015\right)\)

\(=x^2\left(x-1\right)^2+2\left(x-1\right)^2+2013\ge2013\)

Dấu "=" xảy ra khi x - 1 = 0 tức là x = 1

Vậy \(A_{min}=2013\Leftrightarrow x=1\)

x=0 nữa chứ

Bài 2: 

a: Ta có: \(x^2+4x+7\)

\(=x^2+4x+4+3\)

\(=\left(x+2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=-2

\(A=8x^2-4x+\frac{1}{4x^2}+2015\)

\(=\left(4x^2+\frac{1}{4x^2}\right)+\left(4x^2-4x+1\right)+2014\)

\(=\left(4x^2+\frac{1}{4x^2}\right)+\left(2x-1\right)^2+2014\)

Áp dụng bđt AM - GM ta có : \(4x^2+\frac{1}{4x^2}\ge2\sqrt{4x^2.\frac{1}{4x^2}}=2\)

\(\left(2x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(4x^2+\frac{1}{4x^2}\right)+\left(4x^2-4x+1\right)\ge2\)

\(\Rightarrow A=\left(4x^2+\frac{1}{4x^2}\right)+\left(4x^2-4x+1\right)+2014\ge2016\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}4x^2=\frac{1}{4x^2}\\\left(2x-1\right)^2=0\end{cases}}\) \(\Rightarrow x=\frac{1}{2}\)

Vậy \(A_{min}=2016\) tại \(x=\frac{1}{2}\)

\(4x^2+4x+2022=4x^2+4x+1+2021=\left(2x+1\right)^2+2021\ge2021\)

dấu "=" xảy ra \(< =>2x+1=0< =>x=\dfrac{-1}{2}\)

Đặt \(-6x^2+3x+3=0\)

\(\Leftrightarrow-6x^2+6x-3x+3=0\)

\(\Leftrightarrow-6x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)