1.

$x(x+2)(x+4)(x+6)+8$

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

$=a(a+8)+8$ (đặt $x^2+6x=a$)

$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$

Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$

2.

$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$

$=5-(x^2+5x-6)(x^2+5x+6)$

$=5-[(x^2+5x)^2-6^2]$

$=41-(x^2+5x)^2\leq 41$

Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$

a: \(-x^2+2x-4\)

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

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

\(=-\left\lbrack\left(x-1\right)^2+3\right\rbrack=-\left(x-1\right)^2-3\le-3\forall x\)

=>\(\frac{1}{-x^2+2x-4}\ge-\frac13\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(-4x^2+12x-13\)

\(=-\left(4x^2-12x+13\right)\)

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

\(=-\left\lbrack\left(2x-3\right)^2+4\right\rbrack=-\left(2x-3\right)^2-4\le-4\forall x\)

=>\(\frac{12}{-4x^2+12x-13}\ge\frac{12}{-4}=-3\forall x\)

Dấu '=' xảy ra khi 2x-3=0

=>2x=3

=>\(x=\frac32\)

c: Đặt \(A=\frac{x^2-4x-4}{x^2-4x+5}\)

\(=\frac{x^2-4x+5-9}{x^2-4x+5}\)

\(=1-\frac{9}{x^2-4x+5}\)

Ta có: \(x^2-4x+5\)

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

\(=\left(x-2\right)^2+1\ge1\forall x\)

=>\(\frac{9}{\left(x-2\right)^2+1}\le\frac91=9\forall x\)

=>\(-\frac{9}{\left(x-2\right)^2+1}\ge-9\forall x\)

=>\(A=-\frac{9}{\left(x-2\right)^2+1}+1\ge-9+1=-8\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

e: Đặt \(B=\frac{x^2-2011}{4\left(x^2+1\right)}\)

\(=\frac14\cdot\frac{4x^2-8044}{4x^2+4}=\frac14\cdot\frac{x^2-2011}{x^2+1}=\frac14\left(\frac{x^2+1-2012}{x^2+1}\right)=\frac14\left(1-\frac{2012}{x^2+1}\right)\)

Ta có: \(x^2+1\ge1\forall x\)

=>\(\frac{2012}{x^2+1}\le2012\forall x\)

=>\(-\frac{2012}{x^2+1}\ge-2012\forall x\)

=>\(1-\frac{2012}{x^2+1}\ge-2012+1=-2011\forall x\)

=>\(\frac14\left(1-\frac{2012}{x^2+1}\right)\ge-\frac{2011}{4}\forall x\)

Dấu '=' xảy ra khi x=0

 4-3=2( dân chơi mới hiểu)

Chắc là viết thiếu số "1" đấy, sợ lớp 11 còn chưa làm được cơ

Tìm GTNN của biểu thức :

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

Đặt A = \(x^2+2x+4\)

\(\Leftrightarrow A=\left(x^2+2.x.1+1\right)+3\)

\(\Leftrightarrow A=\left(x+1\right)^2+3\)

Ta luôn có : \(\left(x+1\right)^2\ge0\forall x\)

Suy ra : \(\left(x+1\right)^2+3\ge3\forall x\)

Hay A\(\ge3\) với mọi x

Dấu "=" xảy ra khi \(x+1=0\Rightarrow x=-1\)

Nên : \(A_{min}=3khix=-1\)

\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy dấu \("="\) không xảy ra

\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)

\(y_{min}=-3\) khi \(x=1\)

\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)

\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)

\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)

\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)

a) đặt \(A=x^2+x+1\)

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

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

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

Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=-\dfrac{1}{2}\)

b) đặt \(B=2+x-x^2\)

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

\(=-\left(x^2-x-2\right)\)

\(=-\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}-2\right]\)

\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]\)

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

Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\)

Vậy \(MAX_B=\dfrac{9}{4}\) khi \(x=\dfrac{1}{2}\)

c) đặt \(C=x^2-4x+1\)

\(=x^2-2\cdot x\cdot2+2^2-4+1\)

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

Dấu "=" xảy ra khi \(x=2\)

Vậy \(MIN_c=-3\) khi \(x=2\)

d) đặt \(D=4x^2+4x+11\)

\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2-1+11\)

\(=\left(2x+1\right)^2+10\ge10\)

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

Vậy \(MIN_D=10\) khi \(x=-\dfrac{1}{2}\)

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