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

1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)

 \(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)

max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t

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

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

Vậy GTNN cùa A là \(\frac{3}{4}khix=1\)

Ta có:

\(B=\frac{x^4+x^2+5-\frac{19}{20}x^4-\frac{19}{10}x-\frac{19}{20}+\frac{19}{20}\left(x^4+2x^2+1\right)}{x^4+2x^2+1}=\frac{\frac{1}{20}\left(x^4-18x^2+81\right)+\frac{19}{20}\left(x^4+2x^2+1\right)}{x^4+2x^2+1}\)

    \(=\frac{1}{20}.\frac{\left(x^2-9\right)^2}{\left(x^2+1\right)^2}+\frac{19}{20}\ge\frac{19}{20}\)

Vậy GTLN của B là 19/20 khi x = -3 hoăc x = 3.

1. Ta có : \(A=\frac{\left(x+4\right)\left(x+9\right)}{x}=\frac{x^2+13x+36}{x}=x+\frac{36}{x}+13\)

Áp dụng bđt Cauchy : \(x+\frac{36}{x}\ge2\sqrt{x.\frac{36}{x}}=12\)

\(\Rightarrow A\ge25\)

Vậy Min A = 25 \(\Leftrightarrow\begin{cases}x>0\\x=\frac{36}{x}\end{cases}\) \(\Leftrightarrow x=6\)

2. \(B=\frac{\left(x+100\right)^2}{x}=\frac{x^2+200x+100^2}{x}=x+\frac{100^2}{x}+200\)

Áp dụng bđt Cauchy : \(x+\frac{100^2}{x}\ge2\sqrt{x.\frac{100^2}{x}}=200\)

\(\Rightarrow B\ge400\)

Vậy Min B = 400 \(\Leftrightarrow\begin{cases}x>0\\x=\frac{100^2}{x}\end{cases}\) \(\Leftrightarrow x=100\)