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\(A=\dfrac{x\sqrt{x}+x-y+y\sqrt{y}-xy\sqrt{x}-xy\sqrt{y}}{\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x\sqrt{x}\left(1-y\right)+x\left(1-y\sqrt{y}\right)-y\left(1-\sqrt{y}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\left(1-\sqrt{y}\right)\left[x\sqrt{x}\left(1+\sqrt{y}\right)+x+x\sqrt{y}+xy-y\right]}{\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x\sqrt{x}+x\sqrt{xy}+x+x\sqrt{y}+xy-y}{\left(1+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x\left(\sqrt{x}+1\right)+x\sqrt{y}\left(\sqrt{x}+1\right)+y\left(x-1\right)}{\left(1+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x+x\sqrt{y}+y\sqrt{x}-y}{\sqrt{x}+\sqrt{y}}=\sqrt{x}-\sqrt{y}+\sqrt{xy}\)
Để A=2 thì x=2; y=2
\(P=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
\(=\sqrt{xy}+\sqrt{x}-\sqrt{y}\)
Ta có: \(P=\sqrt{xy}+\sqrt{x}-\sqrt{y}=2\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{y}+1\right)=3\)
\(\Rightarrow\left(\sqrt{x}-1,\sqrt{y}+1\right)=\left(1,3;3,1\right)\)
\(\Rightarrow\left(x,y\right)=\left(4,4;16,0\right)\)
\(F=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy}\cdot\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\cdot xy=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
\(=\sqrt{xy}+\sqrt{x}-\sqrt{y}\)
\(P=2\Rightarrow\sqrt{xy}+\sqrt{x}-\sqrt{y}=2\)
\(\Rightarrow\left[{}\begin{matrix}x=y=2\\x=4;y=0\end{matrix}\right.\) (t/m)
làm thế nào để ra được P = \(\sqrt{xy}\)+ \(\sqrt{x}\)- \(\sqrt{y}\) vậy bn ?
a: \(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{x-1}=\dfrac{-2\left(\sqrt{x}-1\right)}{x-1}=\dfrac{-2}{\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}:\left(\dfrac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
c: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
\(P=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{x\left(\sqrt{x}+1\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{x\sqrt{x}+x-y+y\sqrt{y}-yx\sqrt{x}-xy\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{x+y+\sqrt{xy}+\sqrt{x}-\sqrt{y}-xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)+2\sqrt{xy}-xy-1}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=1-\dfrac{\left(\sqrt{xy}-1\right)^2}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}=2\\ \Rightarrow\dfrac{\left(\sqrt{xy}-1\right)^2}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}=1\\ \Leftrightarrow\left(\sqrt{xy}-1\right)^2=\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)\\ \Leftrightarrow xy-2\sqrt{xy}+1=\sqrt{x}-\sqrt{y}+1-\sqrt{xy}\\ \Leftrightarrow\sqrt{x}-\sqrt{y}-xy+\sqrt{xy}=0\)
tự giải quyết tiếp nhá :)) h có việc :)) nếu còn ko bt thì mai làm nốt cho :))