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

$\Rightarrow A\geq \sqrt{10}$ (do $A\geq 0$)

Vậy $A_{\min}=\sqrt{10}$. Giá trị này đạt được khi $(x+4)(6-x)=0\Leftrightarrow x=-4$ hoặc $x=6$

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Áp dụng BĐT Bunhiacopkxy:

$A^2\leq (x+4+6-x)(1+1)=10.2=20$

$\Rightarrow A\leq \sqrt{20}$

Vậy $A_{\max}=\sqrt{20}$

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2

\(A\le\sqrt{\left(3^2+4^2\right)\left(x-1\right)\left(5-x\right)}=10\)

\(A_{max}=10\) khi \(\dfrac{\sqrt{x-1}}{3}=\dfrac{\sqrt{5-x}}{4}\Rightarrow x=\dfrac{61}{25}\)

\(A=3\left(\sqrt{x-1}+\sqrt{5-x}\right)+\sqrt{5-x}\ge3\left(\sqrt{x-1}+\sqrt{5-x}\right)\ge3\sqrt{x-1+5-x}=6\)

\(A_{min}=6\) khi \(x=5\)

+) \(B=6\sqrt{x-2}+6\sqrt{5-x}\Leftrightarrow B^2=\left(6\sqrt{x-2}+6\sqrt{5-x}\right)^2\)

\(=36\left(x-2\right)+36\left(5-x\right)+72\sqrt{\left(x-2\right)\left(5-x\right)}\ge108\Rightarrow B\ge6\sqrt{3}\)

+) \(A=B+2\sqrt{5-x}\ge6\sqrt{3}\)

Vậy \(A_{min}=6\sqrt{3}\)khi x=5

+) Đặt \(a=\sqrt{x-2};b=\sqrt{5-x}\)

+) Ta có: \(a^2+b^2=3\) 

+) \(\left(a^2+b^2\right)\left(6^2+8^2\right)\ge\left(6a+8b\right)^2\Leftrightarrow\left(6a+8b\right)^2\le300\Rightarrow6a+8b\le10\sqrt{3}\)

Dấu = xảy ra khi \(\frac{a}{6}=\frac{b}{8}\Leftrightarrow\frac{\sqrt{x-2}}{6}=\frac{\sqrt{5-x}}{8}\Leftrightarrow\frac{x-2}{36}=\frac{5-x}{64}\Leftrightarrow64x-128=180-36x\Leftrightarrow308=100x\)

\(\Leftrightarrow x=3.08\)

Vậy \(A_{max}=10\sqrt{3}\)khi x=3.08

\(x+y=\sqrt{x+6}+\sqrt{y+6}\ge0\Rightarrow x+y\ge0\)

\(x+y=\sqrt{x+6}+\sqrt{y+6}\le\sqrt{2\left(x+y+12\right)}\)

\(\Rightarrow\left(x+y\right)^2\le2\left(x+y+12\right)\)

\(\Rightarrow\left(x+y+4\right)\left(x+y-6\right)\le0\)

\(\Rightarrow x+y\le6\) (do \(x+y+4>0\))

\(P_{max}=6\) khi \(x=y=3\)

\(x+y=\sqrt{x+6}+\sqrt{y+6}\)

\(\Rightarrow\left(x+y\right)^2=x+y+12+2\sqrt{\left(x+6\right)\left(y+6\right)}\ge x+y+12\)

\(\Rightarrow\left(x+y\right)^2-\left(x+y\right)-12\ge0\)

\(\Rightarrow\left(x+y+3\right)\left(x+y-4\right)\ge0\)

\(\Rightarrow x+y-4\ge0\) (do \(x+y+3>0\))

\(\Rightarrow x+y\ge4\)

\(P_{min}=4\) khi \(\left(x;y\right)=\left(-6;10\right)\) và hoán vị

Ta có: x - \(\sqrt{x+6}\) = \(\sqrt{y+6}\) - y (x; y \(\ge\) -6)

\(\Leftrightarrow\) P = x + y  = \(\sqrt{x+6}+\sqrt{y+6}\)

\(\Leftrightarrow\) P² = x + y + 12 + 2\(\sqrt{\left(x+6\right)\left(y+6\right)}\)

Áp dụng BĐT Cô-si cho 2 số ko âm x + 6 và y + 6 ta có:

\(x+y+12\ge2\sqrt{\left(x+6\right)\left(y+6\right)}\)

\(\Leftrightarrow\) P² \(\le\) x + y + 12 + x + y + 12 = 2x + 2y + 24 = 2P + 24

\(\Leftrightarrow\) P² - 2P - 24 \(\le\) 0

\(\Leftrightarrow\) P² - 36 + 12 - 2P \(\le\) 0

\(\Leftrightarrow\) (P - 6)(P + 6) + 2(6 - P) \(\le\) 0

\(\Leftrightarrow\) (P - 6)(P + 4) \(\le\) 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}P-6\ge0\\P+4\le0\end{matrix}\right.\\\left\{{}\begin{matrix}P-6\le0\\P+4\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}-4\ge P\ge6\left(KTM\right)\\6\ge P\ge-4\left(TM\right)\end{matrix}\right.\)

\(\Rightarrow\) -4 \(\le\) P \(\le\) 6

Vậy ...

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