HOC24
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a, ĐKXĐ: \(\left\{{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)
\(A=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
\(A=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}-\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)
\(A=\sqrt{x^2-1}+1-\left|\sqrt{x^2-1}-1\right|\)
nếu \(\left\{{}\begin{matrix}x\le-\sqrt{2}\\x\ge\sqrt{2}\end{matrix}\right.\) thì \(A=\sqrt{x^2-1}+1-\sqrt{x^2-1}+1=2\) (1)
nếu \(-\sqrt{2}< x< \sqrt{2}\) thì \(A=2\sqrt{x^2-1}\)
vì \(x\ge\sqrt{2}\) thuộc khoảng (1) nên \(A=2\)
\(C=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\right]\)(ĐKXĐ: \(x,y>0\))
\(C=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(C=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\dfrac{1}{xy}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
thay \(\left\{{}\begin{matrix}x=2-\sqrt{3}\\y=2+\sqrt{3}\end{matrix}\right.\) ta có:
\(C=\dfrac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(\sqrt{2}C=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}=-2\)
\(\Rightarrow C=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)
\(Q=\dfrac{\sqrt{x+\sqrt{x^2-y^2}}-\sqrt{x-\sqrt{x^2-y^2}}}{\sqrt{2\left(x-y\right)}}\)
\(Q^2=\dfrac{2x-2\sqrt{\left(x+\sqrt{x^2-y^2}\right)\left(x-\sqrt{x^2-y^2}\right)}}{2\left(x-y\right)}\)
\(Q^2=\dfrac{x-y}{x-y}=1\Rightarrow Q=1\)( do x>y>0)
\(P=-\sqrt{7}+3\sqrt{3}+4\)