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Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
\(A=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
a: \(=\sqrt{3}+1-\sqrt{3}=1\)
b: \(=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c: Sửa đề:\(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{\left(x-1\right)}\)
1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
1) ta có : \(P=\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow P=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(\Leftrightarrow P=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)
2) ta có : \(B=\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}:\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(B=\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}:\left(\dfrac{\sqrt{3}\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}-\dfrac{4}{2\sqrt{6}}+\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\sqrt{3}\sqrt{4+2\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}-4}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}+1\right)^2-4}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+3\right)}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}.\dfrac{2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\sqrt{3}}=\dfrac{1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{6}-\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{\left(\sqrt{6}+\sqrt{2}\right)}{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)
a, \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{15+2\cdot3\cdot\sqrt{6}}-\sqrt{10+2\cdot2\cdot\sqrt{6}}=\sqrt{9+2\cdot3\cdot\sqrt{6}+6}-\sqrt{6+2\cdot\sqrt{6}\cdot2+4}=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}+2\right)^2}=3+\sqrt{6}-\sqrt{6}-2=3-2=1\left(đpcm\right)\)
b, đề không rõ ràng