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 $x=-5$

3.

Đặt $x+3=a; 7-x=b$ thì $a+b=10$ 

$C=a^4+b^4$

Áp dụng BĐT Bunhiacopxky:

$(a^4+b^4)(1+1)\geq (a^2+b^2)^2$

$\Rightarrow C\geq \frac{(a^2+b^2)^2}{2}$
$(a^2+b^2)(1+1)\geq (a+b)^2=100$

$\Rightarrow a^2+b^2\geq 50$

$\Rightarrow C\geq \frac{50^2}{2}=1250$

Vậy $C_{\min}=1250$

Giá trị này đạt tại $a=b=5\Leftrightarrow x=2$

Bài 1:

$\sqrt{x-4}-2$
ĐKXĐ: $x\geq 4$
Ta thấy $\sqrt{x-4}\geq 0$ với mọi $x\geq 4$
$\Rightarrow \sqrt{x-4}-2\geq 0-2=-2$
Vậy gtnn của biểu thức là $-2$. Giá trị này đạt được tại $x-4=0$

$\Leftrightarrow x=4$

Bài 2: $x-\sqrt{x}$

ĐKXĐ: $x\geq 0$

$x-\sqrt{x}=(x-\sqrt{x}+\frac{1}{4})-\frac{1}{4}=(\sqrt{x}-\frac{1}{2})^2-\frac{1}{4}$

$\geq 0-\frac{1}{4}=\frac{-1}{4}$
Vậy gtnn của biểu thức là $\frac{-1}{4}$. Giá trị này đạt được khi $\sqrt{x}-\frac{1}{2}=0$

$\Leftrightarrow x=\frac{1}{4}$

Bài 3:

$x-4\sqrt{x}+10$

ĐKXĐ: $x\geq 0$

Ta có: $x-4\sqrt{x}+10=(x-4\sqrt{x}+4)+6=(\sqrt{x}-2)^2+6\geq 0+6=6$

Vậy gtnn của biểu thức là $6$. Giá trị này đạt được khi $\sqrt{x}-2=0\Leftrightarrow x=4$

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

\(y_{min}=-3\) 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 \(x=-2\)

1.  x≥1 <=> \(\frac{1}{x}\le1\Leftrightarrow\frac{1}{x}+1\le2\Leftrightarrow A\le2\Rightarrow MaxA=2\Leftrightarrow x=1\)

2. Áp dụng bđt cosi cho x>0. ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\Leftrightarrow P\ge2\Rightarrow MinP=2\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x=1\)

3: \(A=\frac{x^2+x+4}{x+1}=\frac{\left(x^2+2x+1\right)-\left(x+1\right)+4}{x+1}=x+1-1+\frac{4}{x+1}\)

áp dụng cosi cho 2 số dương ta có: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)

x=1 nhe nhap minh di ma ket ban voi minh nhe

\(A=\left(x^2+x+1\right)^2-4\left(x+2\right)^2+15\)

\(\Rightarrow A=\left(x^2+x+1\right)^2-\left[2\left(x+2\right)\right]^2+15\)

\(\Rightarrow A=\left(x^2+x+1+2x+2\right)\left(x^2+x+1-2x-2\right)+15\)

\(\Rightarrow A=\left(x^2+3x+3\right)\left(x^2-x-1\right)+15\)

\(\Rightarrow A=\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}+3\right)\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1\right)+15\)

\(\Rightarrow A=\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(1\right)\)

Ta có : \(\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2\ge0,\forall x\\\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\\\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4},\forall x\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[\left(-\dfrac{3}{2}-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[\left(\dfrac{1}{2}+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[4-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[4+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\dfrac{9}{4}+15\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{19}{4}.\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{27}{16}+15\left(x=-\dfrac{3}{2}\right)\\A\ge-\dfrac{95}{16}+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{267}{16}\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)

\(\Rightarrow GTNN\left(A\right)=\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)