tớ ra kết quả là 2+\(\frac{5\sqrt{xy}}{x-\sqrt{xy}+y}\) mà thấy số xấu quá :(

ĐKXĐ:

\(P=\left(\frac{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{x\sqrt{x}-y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{\left(\sqrt{x}+\sqrt{y}\right)}{x-\sqrt{xy}+y}\)

\(=\left(\frac{x\sqrt{y}-y\sqrt{x}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\left(\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\right)\)

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

\(P=\frac{\sqrt{xy}}{x-\sqrt{xy}+\frac{y}{4}+\frac{3y}{4}}=\frac{\sqrt{xy}}{\left(\sqrt{x}-\frac{\sqrt{y}}{2}\right)^2+\frac{3y}{4}}\)

Do \(\left\{{}\begin{matrix}\sqrt{xy}\ge0\\y\ge0\end{matrix}\right.\) \(\Rightarrow P\ge0\) \(\forall x;y\)

a) A = B : C = \(\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right).\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]\). \(\frac{\sqrt{x^3y}+\sqrt{xy^3}}{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}\)

A xác định <=> x > 0 và y > 0

\(B=\left[\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}.\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]=\frac{2}{\sqrt{xy}}+\frac{1}{x}+\frac{1}{y}=\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2\)

\(C=\frac{\sqrt{x}.\left(x+y\right)+\sqrt{y}.\left(x+y\right)}{\sqrt{xy}.\left(x+y\right)}=\frac{\left(\sqrt{x}+\sqrt{y}\right).\left(x+y\right)}{\sqrt{xy}.\left(x+y\right)}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}\)

=> A =  B : C = \(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2\) : \(\left(\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}\right)\) = \(\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}\)

c) \(A=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}\ge2.\sqrt{\frac{1}{\sqrt{y}}.\frac{1}{\sqrt{x}}}=2.\sqrt{\frac{1}{\sqrt{6}}}\)

=> A nhỏ nhất bằng \(2.\sqrt{\frac{1}{\sqrt{6}}}\) khi \(\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{x}}\) => x = y = \(\sqrt{6}\)

Bài 2 :

a) \(ĐKXĐ:\hept{\begin{cases}x;y>0\\x\ne y\end{cases}}\)

b) \(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\right):\frac{x\sqrt{xy}+y\sqrt{xy}}{\sqrt{xy}\left(y-x\right)}\)

\(\Leftrightarrow A=\frac{x-\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}:\frac{x+y}{y-x}\)

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

\(\Leftrightarrow A=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(y-x\right)}{x+y}\)

c) Thay \(x=4+2\sqrt{3},y=4-2\sqrt{3}\)vào A, ta được :

   \(A=\frac{\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\left(4-2\sqrt{3}-4-2\sqrt{3}\right)}{4+2\sqrt{3}+4-2\sqrt{3}}\)

\(\Leftrightarrow A=\frac{\left(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\right).\left(-4\sqrt{3}\right)}{8}\)

\(\Leftrightarrow A=\frac{\left(1+\sqrt{3}-\sqrt{3}+1\right).\left(-4\sqrt{3}\right)}{8}=\frac{-8\sqrt{3}}{8}=-\sqrt{3}\)

Vậy ....

Bài 1:

\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}=\frac{2\sqrt{2\cdot4}-\sqrt{3\cdot4}}{\sqrt{2\cdot9}-\sqrt{16\cdot3}}-\frac{\sqrt{5}+\sqrt{9\cdot3}}{\sqrt{30}-\sqrt{2}}\)

\(=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}-\sqrt{2}}=\frac{\left(4\sqrt{2}-2\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)-\left(\sqrt{5}+3\sqrt{3}\right)\left(3\sqrt{2}-4\sqrt{3}\right)}{\left(3\sqrt{2}-4\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)}\)

\(=\frac{4\sqrt{60}-8-2\sqrt{90}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{3\sqrt{60}-6-4\sqrt{90}+4\sqrt{6}}\)

\(=\frac{8\sqrt{15}-8-6\sqrt{10}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{6\sqrt{15}-6-12\sqrt{10}+4\sqrt{6}}\)

\(=\frac{12\sqrt{15}-2\sqrt{10}-7\sqrt{6}+28}{6\sqrt{15}-12\sqrt{10}+4\sqrt{6}-6}\)