A=\(x^2+\left(3y\right)^2+4-6xy+12y-4x+x^2+4x+4+1996\)

A=\(\left(x-3y-2\right)^2+\left(x+2\right)^2+1996\ge1996\)

Vay gtnn cua A la 1996.Dat duoc khi va chi khi \(\hept{\begin{cases}x+2=0\\x-3y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\3y=-2-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-2\\y=-\frac{4}{3}\end{cases}}}\)

Study well

\(M=\left|2x-3\right|+\frac{\left|4x-1\right|}{2}\Rightarrow2M=\left|4x-6\right|+\left|4x-1\right|\)

Áp dụng bất đẳng thức : \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) . Dấu đẳng thức xảy ra khi a,b cùng dấu.

Được : \(2M=\left|6-4x\right|+\left|4x-1\right|\ge\left|6-4x+4x-1\right|=5\) \(\Rightarrow2M\ge5\)

\(\Rightarrow M\ge\frac{5}{2}\) . Dấu đẳng thức xảy ra \(\Leftrightarrow\begin{cases}6-4x\ge0\\4x-1\ge0\end{cases}\)\(\Leftrightarrow\frac{1}{4}\le x\le\frac{3}{2}\)

Vậy Min M = \(\frac{5}{2}\Leftrightarrow\frac{1}{4}\le x\le\frac{3}{2}\)

\(M=5x^2+y^2-2x+2y+2xy+2004\)

\(=\left(x^2+2x+1\right)+2y\left(x+1\right)+y^2+4x^2-4x+1+2002\)

\(=\left(x+1\right)^2+2y\left(x+1\right)+y^2+\left(2x-1\right)^2+2002\)

\(=\left(x+1+y\right)^2+\left(2x-1\right)^2+2003\ge2002\) với mọi x,y

=> \(M_{min}=2002\Leftrightarrow\left\{{}\begin{matrix}x+y+1=0\\2x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(M_{min}=2002\)

a)x²-2x+m= (x-1)²+m-1 \(\ge m-1\) Min =2 => m-1 = 2 <=> m = 3

b) = 4x²-2x+6x+m= 4x²+4x+m = (2x+1)²+m-1 \(\ge m-1\) Min=1998 <=> m-1 = 1998 <=> m = 1999

a, Ta có: \(\left(x-1\right)^4\ge0\forall x\)

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

\(\Rightarrow M=\left(x-1\right)^4+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

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

Vậy \(M_{min}=\dfrac{1}{4}\Leftrightarrow x=1\)

b, Ta có: \(\left(2x-1\right)^2\ge0\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

\(\left|y-1\right|\ge0\forall y\)

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

\(\Rightarrow N=3+\left(2x-1\right)^2+\left|y-1\right|\ge3\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)

Vậy \(N_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)