Bạn cần làm gì với biểu thức này?
 

B

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

\(F=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy}\cdot\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\cdot xy=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

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

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

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

\(\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

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

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

Ta có: \(C=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

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

b: \(\left(\sqrt{x}+\sqrt{y}\right)^2=\left(\sqrt{2-\sqrt3}+\sqrt{2+\sqrt3}\right)^2\)

\(=2-\sqrt3+2+\sqrt3+2\cdot\sqrt{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}=4+2=6\)

\(\sqrt{xy}=\sqrt{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}=\sqrt{4-3}=1\)

\(x+y=2+\sqrt3+2-\sqrt3=4\)

\(\sqrt{x}-\sqrt{y}=\sqrt{2-\sqrt3}-\sqrt{2+\sqrt3}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{4-2\sqrt3}-\sqrt{4+2\sqrt3}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3-1\right)^2}-\sqrt{\left(\sqrt3+1\right)^2}\right)=\frac{1}{\sqrt2}\left(\sqrt3-1-\sqrt3-1\right)=-\frac{2}{\sqrt2}=-\sqrt2\)

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

\(=\frac{2\cdot4-\sqrt2}{1\cdot4\cdot6}=\frac{8-\sqrt2}{24}\)

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