x² + y² = \(\sqrt{9-4\sqrt{5}}+\sqrt{14-6\sqrt{5}}\) = \(\sqrt{5}-2+3-\sqrt{5}=1\)

Ta có 

P = xy \(\le\frac{x^2+y^2}{2}=\frac{1}{2}\)

M= \(\sqrt{x}-1+\sqrt{y-x}\)(đk : \(y\ge x\ge0\)

Áp dụng bđt cosi vs hai số dương có:

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

\(\sqrt{y-x}\le\frac{y-x+1}{2}\)

=> \(\sqrt{x}+\sqrt{y-x}\le\frac{x+1}{2}+\frac{y-x+1}{2}\) <=> \(\sqrt{x}-1+\sqrt{y-x}\le\frac{x}{2}+\frac{1}{2}+\frac{y}{2}-\frac{x}{2}+\frac{1}{2}-1\)

<=> \(M\le\frac{y}{2}\) (1)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}1=y-x\\x=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

Từ (1) => M=1 tại y=2,x=1

Vậy maxM=1 <=> x=1,y=2

\(A=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)(đk: \(x,y>0\) ,\(x\ne y\))

=\(\frac{x+y+2\sqrt{xy}-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\) (vì x,y>0)

=\(\frac{x-2\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}+\sqrt{x}-\sqrt{y}\)

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

Vậy A= \(2\sqrt{x}-2\sqrt{y}\)

\(a,B=\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{1-xy}\right):\left(\frac{1-xy+x+y+2xy}{1-xy}\right)\)

\(B=\frac{\sqrt{x}+\sqrt{y}+x\sqrt{y}+y\sqrt{x}+\sqrt{x}-\sqrt{y}-x\sqrt{y}+y\sqrt{x}}{1-xy}.\frac{1-xy}{1+xy+x+y}\)

\(B=\frac{2\sqrt{x}+2y\sqrt{x}}{x\left(y+1\right)+\left(y+1\right)}\)

\(B=\frac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}\)

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

\(b,B=\frac{2\sqrt{\frac{2}{2+\sqrt{3}}}}{\frac{2}{2+\sqrt{3}}+1}\)

\(\frac{2\sqrt{\frac{4}{4+2\sqrt{3}}}}{\frac{4}{4+2\sqrt{3}}+1}\)

\(B=\frac{2\sqrt{\frac{4}{\left(\sqrt{3}+1\right)^2}}}{\frac{4}{\left(\sqrt{3}+1\right)^2}+1}\)

\(B=\frac{2.2}{\sqrt{3}+1}:\frac{4+2\sqrt{3}}{\sqrt{3}+1}\)

\(B=\frac{4}{\left(\sqrt{3}+1\right)^2}\)

\(B=\left(\frac{2}{\sqrt{3}+1}\right)^2\)

\(c,B=\frac{2\sqrt{x}}{x+1}\)

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

ta có :

\(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\)

dấu "=" xảy ra khi \(x=1\)

\(< =>MAX:B=\frac{2}{2}=1\)

Đk: x \(\ge\)0; y \(\ge\)0; xy \(\ne\)1

Ta có: B = \(\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\frac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\frac{x+y+2xy}{1-xy}\right)\)

B = \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}+1\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\frac{1-xy+x+y+2xy}{1-xy}\)

B = \(\frac{x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}\cdot\frac{1-xy}{x+y+xy+1}\)

B = \(\frac{2\sqrt{x}+2y\sqrt{x}}{\left(y+1\right)\left(x+1\right)}=\frac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\frac{2\sqrt{x}}{x+1}\)

b) Ta có: \(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4-2\sqrt{3}}{4-3}=4-2\sqrt{3}\)

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

Do đó, B = \(\frac{2.\left(\sqrt{3}-1\right)}{4-2\sqrt{3}+1}=\frac{2\sqrt{3}-2}{5-2\sqrt{3}}=\frac{\left(2\sqrt{3}-2\right)\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}=\frac{10\sqrt{3}+12-10-4\sqrt{3}}{25-12}\)

B = \(\frac{6\sqrt{3}+2}{13}\)

c) Ta có: \(\frac{1}{B}=\frac{x+1}{2\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\ge2\cdot\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{2\sqrt{x}}}=2\cdot\sqrt{\frac{1}{4}}=1\)(đk: x \(\ne\)0)

=> \(B\le\frac{1}{1}=1\)Dấu "==" xảy ra<=> \(\frac{\sqrt{x}}{2}=\frac{1}{2\sqrt{x}}\) => \(2\sqrt{x}=2\) => \(x=1\)

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