\(a,x-9+y-2\sqrt{xy}\left(x;y>0\right)\)

\(=\left(\sqrt{x}\right)^2-2\sqrt{x}\sqrt{y}+\left(\sqrt{y}\right)^2-9\)

\(=\left(\sqrt{x}-\sqrt{y}\right)^2-9\)

\(=\left(\sqrt{x}-\sqrt{y}+3\right)\left(\sqrt{x}-\sqrt{y}-3\right)\)

\(b,\text{ đkxđ }x\ge0\)

\(x-5\sqrt{x}+6=\left(\sqrt{x}\right)^2-2\sqrt{x}-3\sqrt{x}+6\)

\(=\sqrt{x}.\left(\sqrt{x}-2\right)-3.\left(\sqrt{x}-2\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)

\(c,đ\text{kxđ }x\ge0\)

\(x-2\sqrt{x}-3=\left(\sqrt{x}\right)^2+\sqrt{x}-3\sqrt{x}-3\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)+3.\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\)

\(d,\text{đkxđ }x\ge0\)

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

\(=\sqrt{x}.\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)\)

\(B=2\sqrt{2016}=\sqrt{2016}+\sqrt{2016}>\sqrt{2014}+\sqrt{2016}=A\)

 Vậy A < B

ta có \(B=2\sqrt{2016}=\sqrt{2016}+\sqrt{2016}>\sqrt{2014}+\sqrt{2016}=A\)

Vậy A<B

a/ \(=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)

câu b k bik lm ^^

b/ \(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+2}=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)

Điều kiện: x >= 0

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

Vì \(\sqrt{x}+1\ge1\)=> \(\frac{3}{\sqrt{x}+1}\le\frac{3}{1}=3\) => \(-\frac{3}{\sqrt{x}+1}\ge-3\) => P \(\ge\) 2 - 3 = -1

Vậy GTNN ciuar P = -1 khi x = 0