ĐK: x > = 3

pt <=> \(x^2-5x+4+\left(\sqrt{2x+1}-3\right)+\left(\sqrt{x-3}-1\right)=0\)

<=> \(\left(x-1\right)\left(x-4\right)+\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{x-3}+1}=0\)

<=> \(\left(x-4\right)\left(\left(x-1\right)+\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}\right)=0\)

<=> x - 4 = 0  vì \(\left(x-1\right)+\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}>0;\forall x\ge3\)

<=> x = 4  tm 

Vậy:...

 \(t=\sqrt{2x-3}=>\frac{t^2+3}{2}=x\)

\(=>P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}=\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\)

ta có \(\frac{\left(t-2\right)^2}{2}\ge0\left(\forall t\right)\)

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

minP=-1/2

dấu = xảy ra khi x=7/2

a) \(t=\sqrt{2x-3}\ge0\)

<=> \(t^2=2x-3\)

<=> \(x=\frac{t^2+3}{2}\)

=> \(P=\frac{t^2+3}{2}-2t\)

b) khi đó: \(P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}\ge-\frac{1}{2}\)

Dấu "=" xảy ra <=> t = 2  khi đó: x = 7/2

\(P=\sqrt{3a^2+2ab+3b^2}+...+\sqrt{3c^2+2ac+3a^2}\)

ta có:

\(\sqrt{3a^2+2ab+3b^2}\ge\sqrt{2}\left(a+b\right)\Leftrightarrow3a^2+2ab+3b^2\ge2a^2+4ab+2b^2\Leftrightarrow\left(a-b\right)^2\ge0\left(đ\right)\)

\(\text{tương tự suy ra:}P\ge2\sqrt{2}\left(a+b+c\right)\)

\(\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=4\Leftrightarrow\sqrt{ab}+\sqrt{a}+\sqrt{b}=3\)

\(\text{Ta có:}M\ge a+b\Rightarrow2M+2\ge a+b+a+1+b+1\ge2\left(\sqrt{ab}+\sqrt{a}+\sqrt{b}\right)\left(\text{theo cô si}\right)=6\)

\(\Rightarrow M\ge2\left(\text{dấu "=" xảy ra khi:}a=b=1\right)\)

Ta có: \(3=x^2+y^2+z^2\ge xy+yz+xz\ge\frac{\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}{3}\)

=> \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le3\)

\(M=\frac{xyz}{x^2+yz}+\frac{xyz}{y^2+zx}+\frac{xyz}{z^2+xy}\)

\(\le\frac{xyz}{2x\sqrt{yz}}+\frac{xyz}{2y\sqrt{xz}}+\frac{xyz}{2z\sqrt{xy}}\)

\(=\frac{1}{2}\left(\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\right)\le\frac{3}{2}\)

Dấu "=" xảy ra <=> x = y = z=1

Ta có: 

\(P=\frac{x+12}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}\)

\(=\left(\sqrt{x}+2\right)+\frac{16}{\sqrt{x}+2}-4\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=4\)

Dấu "=" xảy ra <=> \(\sqrt{x}+2=\frac{16}{\sqrt{x}+2}\Leftrightarrow\sqrt{x}+2=4\Leftrightarrow x=4\) thỏa mãn

=> min P = 4 tại x = 4.