\(A=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(x^2-4x+4\right)-3\)

\(A=\left(x-2y+1\right)^2+\left(x-2\right)^2-3\ge-3\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;\dfrac{3}{2}\right)\)

a) \(M=x^2-3x+10\)

\(M=x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}+\dfrac{31}{4}\)

\(M=\left(x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}\right)+\dfrac{31}{4}\)

\(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\)

Mà: \(\left(x-\dfrac{3}{2}\right)^2\ge0\) nên: \(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)

Dấu "=" xảy ra 

\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=\dfrac{31}{4}\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

Vậy: \(M_{min}=\dfrac{31}{4}\) với \(x=\dfrac{3}{2}\)

b) \(N=2x^2+5y^2+4xy+8x-4y-100\)

\(N=x^2+x^2+4y^2+y^2+4xy+8x-4y-120+16+4\)

\(N=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-120\)

\(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\)

Mà:

\(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) nên \(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\ge120\)

Dấu "=" xảy ra:

\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-4+2y=0\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)

Vậy: \(N_{min}=120\) khi \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)

a

\(M=x^2-3x+10=x^2-2.\dfrac{3}{2}.x+\dfrac{9}{4}+\dfrac{31}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)

Min M \(=\dfrac{31}{4}\) khi và chỉ khi \(x=\dfrac{3}{2}\)

b

\(N=2x^2+5y^2+4xy+8x-4y-100\\ =x^2+8x+16+y^2-4y+4+x^2+4xy+4y^2-120\\ =\left(x+4\right)^2+\left(y-2\right)^2+\left(x+2y\right)^2-120\ge-120\)

Min N \(=-120\) khi và chỉ khi \(x=-4\) và \(y=2\)

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

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

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

b: Ta có: \(4x^2-3x+10\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

a: Ta có: \(A=x^2+3x+4\)

\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

\(A=3x^2+4xy+4y^2-3x-2y+15\)

\(=\left(x^2+4xy+4y^2\right)-\left(x+2y\right)+\frac{1}{4}+2x^2-2x+\frac{59}{4}\)

\(=\left(x+2y-\frac{1}{2}\right)^2+2\left(x-\frac{1}{2}\right)^2+\frac{57}{4}\ge\frac{57}{4}\)

Đẳng thức xảy ra khi x =1/2; y =0

Vậy..