1)\(2x^2+9y^2-6xy-6x-12y+2004\)

\(=x^2+x^2-6xy+9y^2-6x-12y+2004\)

\(=x^2+\left(x-3y\right)^2-10x+4x-12y+2004\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+2004\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+4+25+1975\)

\(=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x^2-10x+25\right)+1975\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\)

Dấu "=" khi \(\begin{cases}\left(x-5\right)^2=0\\\left(x-3y+2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)

Vậy Min=1975 khi \(\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)

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

Đặt \(t=x^2+x\) ta có:

\(t\left(t-4\right)=t^2-4t+4-4\)

\(=\left(t-2\right)^2-4\ge-4\)

Dấu "=" khi \(t-2=0\Leftrightarrow t=2\Leftrightarrow x^2+x=2\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)

Vậy Min=-4 khi \(\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)

3)\(\left(x^2+5x+5\right)\left[\left(x+2\right)\left(x+3\right)+1\right]\)

\(=\left(x^2+5x+5\right)\left[x^2+5x+6+1\right]\)

Đặt \(t=x^2+5x+5\) ta có:

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

Dấu "=" khi \(t+\frac{1}{2}=0\Leftrightarrow t=-\frac{1}{2}\Leftrightarrow x^2+5x+5=-\frac{1}{2}\)\(\Leftrightarrow x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)

Vậy Min=\(-\frac{1}{4}\) khi \(x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)

4)\(\left(x-1\right)\left(x-3\right)\left(x^2-4x+5\right)\)

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

Đặt \(t=x^2-4x+3\) ta có:

\(t\left(t+2\right)=t^2+2t+1-1=\left(t+1\right)^2-1\ge-1\)

Dấu "=" khi \(t+1=0\Leftrightarrow t=-1\Leftrightarrow x^2-4x+3=-1\Leftrightarrow x=2\)

Vậy Min=-1 khi x=2

Thank you !

Nhóm (x+1)(x+4)=t

(x+2)(x+3)=t+2

A=t(t+2)+5

A=t²+2t+5

A=(t+1)²+4

Min_A=4 khi ............

Ta có : E = (x - 1) (x + 2)(x + 3)(x + 6)

=> E = [(x - 1)(x + 6)][(x + 2)(x + 3)]

=> E = (x² + 5x - 6)(x² + 5x + 6)

=> E = (x² + 5x)² - 6²

=> E = (x² + 5x)² - 36

Mà : (x² + 5x)² \(\ge0\forall x\)

Nên : (x² + 5x)² - 36 \(\ge-36\forall x\)

Vậy GTNN của biểu thức là 36 tại x² + 5x = 0 => x(x + 5) = 0 => x = 0 ; -5 

Bài 1:

\(A=x^2-6x+12=\left(x^2-6x+9\right)+3=\left(x-3\right)^2+3\ge3\)Vậy minA = 3 khi x = 3

\(B=x^2-4x+15=\left(x^2-4x+4\right)+11=\left(x-2\right)^2+11\ge11\)Vậy minB = 11 khi x = 2

\(C=x^2+y^2-2x+6y+17=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)+7\)\(=\left(x-1\right)^2+\left(y+3\right)^2+7\ge7\)Vậy minC = 7 khi x = 1, y = -3

Bài 2:

Câu M mình thấy nó hơi lạ, nếu đề là \(M=4x-x^2+10\)thì mình giải được, bạn xem lại nhé!

\(M=4x-x^2+10=-\left(x^2-4x+4\right)+14=-\left(x-2\right)^2+14\le14\)Vậy maxM = 14 khi x = 2

\(N=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)Vậy maxN = \(\frac{1}{4}\)khi x = \(\frac{1}{2}\).