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

\(M=\left[\frac{\sqrt{x}\left(2\sqrt{x}+3\right)}{2x+2\sqrt{x}+3\sqrt{x}+3}+\frac{2}{\sqrt{x}+1}\right].\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\left[\frac{\sqrt{x}\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}+\frac{2}{\sqrt{x}+1}\right].\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}.\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2018}{\sqrt{x}+1}\)

\(\frac{\sqrt{x}+2018}{\sqrt{x}+1}=1+\frac{2017}{\sqrt{x}+1}\le2018\)

Dấu "=" xảy ra \(\Leftrightarrow\)

... 

1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)

Đạt được khi x = 9

2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)

\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)

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

Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)

Không có GTLN nhé

a: ĐKXĐ: x>=0

Ta có: \(x-\sqrt{x}+1\)

\(=x-\sqrt{x}+\frac14+\frac34=\left(\sqrt{x}-\frac12\right)^2+\frac34\ge\frac34\forall x\) thỏa mãn ĐKXĐ

=>\(A=\frac{1}{x-\sqrt{x}+1}\le1:\frac34=\frac43\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{x}-\frac12=0\)

=>\(\sqrt{x}=\frac12\)

=>\(x=\frac14\left(nhận\right)\)

b:

ĐKXĐ: x>=0

\(B=\frac{2x-2\sqrt{x}+5}{x-\sqrt{x}+2}\)

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

Ta có: \(x-\sqrt{x}+2\)

\(=x-\sqrt{x}+\frac14+\frac74=\left(\sqrt{x}-\frac12\right)^2+\frac74\ge\frac74\forall x\) thỏa mãn ĐKXĐ

=>\(\frac{1}{x-\sqrt{x}+2}\le1:\frac74=\frac47\forall x\) thỏa mãn ĐKXĐ

=>\(\frac{1}{x-\sqrt{x}+2}+2\le\frac74+2=\frac{15}{4}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{x}-\frac12=0\)

=>\(\sqrt{x}=\frac12\)

=>\(x=\frac14\) (nhận)