a,\(A=\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=\left(x^2+6x+5\right)\left(x^2+6x+8\right)\)

đặt \(x^2+6x+5=t=>t\left(t+3\right)=t^2+3t=t^2+2.\dfrac{3}{2}t+\dfrac{9}{4}-\dfrac{9}{4}\)

\(=\left(t+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}< =>t=\dfrac{-3}{2}\)

\(=>A\)\(=-\dfrac{3}{2}\left(-\dfrac{3}{2}+3\right)=-2,25\)

Vậy Min A\(=-2,25\)

b,\(B=-x^2-4x-9y^2-6y-6\)

\(=-\left(x^2+4x+4\right)-\left(3y\right)^2-2.3y-1-1\)

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

dấu"=' xảy ra\(< =>x=-2,y=-\dfrac{1}{3}\)

a.

$(x+1)(x+2)(x+4)(x+5)=(x+1)(x+5)(x+2)(x+4)=(x^2+6x+5)(x^2+6x+8)$

$=a(a+3)$ với $a=x^2+6x+5$

$=a^2+3a=(a^2+3a+\frac{9}{4})-\frac{9}{4}$

$=(a+\frac{3}{2})^2-\frac{9}{4}$

$=(x^2+6x+\frac{13}{2})^2-\frac{9}{4}\geq \frac{-9}{4}$

Vậy gtnn của biểu thức là $\frac{-9}{4}$. Giá trị này đạt tại $x^2+6x+\frac{13}{2}=0$

$\Leftrightarrow x=\frac{-6\pm \sqrt{10}}{2}$

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

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

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

a/ \(A=x^2+y^2-2x+6y+12\)

\(=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)+2\)

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

Với mọi x, y ta có :

\(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+3\right)^2\ge0\)

\(\Leftrightarrow A\ge3\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

Vậy....

b/ \(B=-4x^2-9y^2-4x+6y+3\)

\(=-\left(4x^2+4x+1\right)-\left(9y^2+6y+1\right)+1\)

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

Với mọi x, y ta có :

\(\left\{{}\begin{matrix}\left(2x+1\right)^2\ge0\\\left(3y+1\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\left(2x+1\right)^2\le0\\-\left(3y+1\right)^2\le0\end{matrix}\right.\)

\(\Leftrightarrow-\left(2x+1\right)^2-\left(3y+1\right)^2\le0\)

\(\Leftrightarrow B\le1\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=-\frac{1}{3}\end{matrix}\right.\)

\(a)\)

\(A=2x^2+x\)

\(\Leftrightarrow A=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)

\(MinA=\frac{-1}{8}\)khi \(x=\frac{-1}{4}\)

\(b)\)

\(B=x^2+2x+y^2-4y+6\)

\(\Leftrightarrow B=x^2+2x+1+y^2-4y+4+1\)

\(\Leftrightarrow B=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\)

Dấu '' = '' xảy ra khi: \(x=-1;y=2\)

\(c)\)

\(C=4x^2+4x+9y^2-6y-5\)

\(\Leftrightarrow C=4x^2+4x+1+9y^2-6y+1-7\)

\(\Leftrightarrow C=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)

Dấu '' = '' xáy ra khi: \(x=\frac{-1}{2};y=\frac{1}{3}\)

Giúp mình với ạ,cảm ơn mọi người

b: Ta có: \(B=x^2+4x+9y^2-6y-1\)

\(=x^2+4x+4+9y^2-6y+1-6\)

\(=\left(x+2\right)^2+\left(3y-1\right)^2-6\ge-6\forall x,y\)

Dấu '=' xảy ra khi x=-2 và \(y=\dfrac{1}{3}\)

a) \(C=4x^2+3y^2+4xy-4x-10y+7=\left[4x^2+4x\left(y-1\right)+\left(y-1\right)^2\right]+2\left(y^2-4y+4\right)-2=\left(2x+y-1\right)^2+2\left(y-2\right)^2-2\ge-2\)

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

d) \(D=x^2-2xy+6y^2-12x+2y+45=\left[x^2-2x\left(y+6\right)+\left(y+6\right)^2\right]+5\left(y^2-2y+1\right)+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

\(minD=4\Leftrightarrow\) \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

a)

\(A=x^2+7x+8=x^2+2.x.\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2-\dfrac{17}{4}\\ =\left(x+\dfrac{7}{2}\right)^2-\dfrac{17}{4}\ge-\dfrac{17}{4}\)

\(a,-x^2+2x+5=-\left(x^2-2x-5\right)=-\left(x^2-2x+1-6\right)=-\left(x-1\right)^2+6\le6\)

dấu'=' xảy ra<=>x=1=>Max A=6

\(b,B=-x^2-y^2+4x+4y+2=-x^2+4x-4-y^2+4x-4+10\)

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

\(=-\left(x-2\right)^2-\left(y-2\right)^2+10=-\left[\left(x-2\right)^2+\left(y-2\right)^2\right]+10\le10\)

dấu"=" xảy ra<=>x=y=2=>Max B=10

\(c,C=x^2+y^2-2x+6y+12=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

dấu'=' xảy ra<=>x=1,y=-3=>MinC=2

\(a,-x^3+\left(x-3\right)\left[\left(2x+1\right)^2-2\left(\dfrac{3}{2}x^2+\dfrac{1}{2}x-4\right)\right]\\ =-x^3+\left(x-3\right)\left(4x^2+4x+1-3x^2-x+8\right)\\ =-x^3+\left(x-3\right)\left(x^2+3x+9\right)\\ =-x^3+\left(x^3-27\right)=-27\)

\(b,\left(x+2y\right)^3-\left(x-3y\right)\left(x^2+3xy+9y^2\right)-6y\left(x^2+2xy-\dfrac{35}{6}y^2\right)\\ =x^3+6x^2y+12xy^2+8y^3-x^3+27y^3-6x^2y-12xy^2+35y^3\\ =0\)

ta có : \(C=x^2-2x+9y^2-6y+3=\left(x^2-2x+1\right)+\left(9y^2-6y+1\right)+1\)

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

ta có : \(\left(x-1\right)^2\ge0\) với mọi \(x\) và \(\left(3y-1\right)^2\ge0\) với mọi \(y\)

\(\Rightarrow C=\left(x-1\right)^2+\left(3y-1\right)^2+1\ge1\) với mọi \(x;y\)

\(\Rightarrow\) giá trị nhỏ nhất của \(C\) là \(1\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(3y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\3y-1=0\end{matrix}\right.\)

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

vậy \(GTNN\) của \(C=x^2-2x+9y^2-6y+3\) là \(1\) khi \(x=1;y=\dfrac{1}{3}\)