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 $a=b=5\Leftrightarrow x=2$

b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:

\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)

\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)

\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)

(x-1)(x+2)(x+3)(x+6) 
=[(x-1)(x+6)][(x+2)(x+3)] 
=(x^2+5x-6)(x^2+5x+6) 
=(x^2+5x)^2-36>=-36 
=>min=-36<=>x=0 hoặc x=-5

\(VìA=(x-1).(x+2).(x+3).(x+6)\)\(\Rightarrow\)\(A=x.(-1+2+3+6)\)\(\Rightarrow\)\(A=x.10\)

Vì A nhỏ nhất \(\Rightarrow\)A=0 mà A=x.10\(\Rightarrow\)0=x.10\(\Rightarrow\)x=0\(:\)10\(\Rightarrow\)x=0

\(Vậy\) \(A\) \(nhỏ\) \(nhất\) \(khi\) x=0

\(Vì\)\(A=\) \((x-1).(x+2).(x+3).(x+6)\)\(\Rightarrow\)\(A= \)\(x.(-1+2+3+6)\)\(\Rightarrow\)\(A=x.10\)

\(Vì\) \(A\)  \(nhỏ\)  \(nhất\)\(\Rightarrow\)\(A=\)0 \(mà\) \(A=x.10\) \(\Rightarrow\)\(x.10= 0\)\(\Rightarrow\)\(x=0:10\)\(\Rightarrow\)\(x=0\)

\(Vậy\) \(A \)  \(nhỏ\)  \(nhất\) \(khi\)  \(x=0\)

A=(x-1)(x+2)(x+3)(x+6)+12

=(x-1)(x+6)(x+2)(x+3)+12

=(x²+6x-x-6)(x²+3x+2x+6)+12

=(x²+5x)²-6.6+12

=(x²+5x)²-36+12

=(x²+5x)²-24

\(\Rightarrow\) (x²+5x)²\(\ge\)0 voi moi x

\(\Rightarrow\) (x²+5x)²\(\ge\) -24

Vay GTNN la -24

Dấu "=" xảy ra khi : x²+5x=0

                             x(x+5)=0

                             =>x=0 va -5 

Nhớ k nha

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)