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a) +) Điều kiện : x \(\ge\) 0 ; y \(\ge\) 0 ; y \(\ne\) 1 ; x; y không đồng thời bằng 0
+) \(P=\frac{x\left(\sqrt{x}+1\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{x\sqrt{x}+x-y+y\sqrt{y}-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y-\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+y-\sqrt{xy}+\sqrt{x}-\sqrt{y}-xy}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(x+\sqrt{x}\right)+\left(y-xy\right)-\left(\sqrt{xy}+\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(1+\sqrt{x}\right)\sqrt{x}+y\left(1-x\right)-\sqrt{y}\left(\sqrt{x}+1\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(1+\sqrt{x}\right)\left(\sqrt{x}+y-y\sqrt{x}-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-y\sqrt{x}\right)+\left(y-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)}=\frac{\sqrt{x}\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)-\sqrt{y}\left(1-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)}\)
\(P=\sqrt{x}\left(1+\sqrt{y}\right)-\sqrt{y}=\sqrt{x}-\sqrt{y}+\sqrt{xy}\)
b) Để P = 2 <=> \(\sqrt{x}-\sqrt{y}+\sqrt{xy}=2\) <=> \(\sqrt{x}+\sqrt{xy}=\sqrt{y}+2\)
<=> \(\left(\sqrt{x}+\sqrt{xy}\right)^2=\left(\sqrt{y}+2\right)^2\)
<=> \(x+xy+2x\sqrt{y}=y+4+4\sqrt{y}\)
<=> \(x+xy-y+\left(2x-4\right)\sqrt{y}=4\)(*)
P = 2 <=> (x; y) thỏa mãn (*)
a) ĐKXĐ : \(x,y\ge0;y\ne1;x+y\ne0\)
\(P=\frac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\frac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\frac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{x\left(1+\sqrt{x}\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{\left(x-y\right)+\left(x\sqrt{x}+y\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+x-\sqrt{xy}+y-xy\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-\sqrt{y}\left(\sqrt{x}+1\right)+y\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1+\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}-\sqrt{y}+x-y\sqrt{x}}{1-\sqrt{y}}=\frac{\sqrt{x}\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)-\sqrt{y}\left(1-\sqrt{y}\right)}{1-\sqrt{y}}\)
\(=\sqrt{x}+\sqrt{xy}+\sqrt{y}\)
Vậy P \(=\sqrt{x}+\sqrt{xy}+\sqrt{y}\)
b) ĐKXĐ : \(x,y\ge0;y\ne1;x+y\ne0\)
\(P=2\Leftrightarrow\) \(\sqrt{x}+\sqrt{xy}+\sqrt{y}=2\) ( * )
\(\Leftrightarrow\sqrt{x}\left(1+\sqrt{y}\right)-\left(\sqrt{y}+1\right)=1\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{y}+1\right)=1\)
Có : \(1+\sqrt{y}\ge1\Rightarrow\sqrt{x}-1\le1\Leftrightarrow0\le x\le4\Rightarrow x=0;1;2;3;4\)
Thay x = 0 ; 1 ; 2 ; 3 ;4 vào ( * )
Ta có các cặp giá trị : x =4 ; y = 0 và x = 2 ; y = 2 ( TM )
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)