\(C=\sqrt{\dfrac{x-6\sqrt{x}+9}{x+6\sqrt{x}+9}}=\sqrt{\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)^2}}=\dfrac{\left|\sqrt{x}-3\right|}{\sqrt{x}+3}\)
Vì \(x\ge9\Rightarrow\sqrt{x}\ge3\Leftrightarrow\sqrt{x}-3\ge0\)
\(\Rightarrow C=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(D=\dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\) (\(x;y\ne1;y\ge0\))
\(=\dfrac{x-1}{\sqrt{y}-1}.\dfrac{\sqrt{\left(\sqrt{y}-1\right)^2}}{\left(x-1\right)^2}=\dfrac{\left|\sqrt{y}-1\right|}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
TH1: \(\sqrt{y}-1>0\Leftrightarrow y>1\)
\(\Rightarrow D=\dfrac{\sqrt{y}-1}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
TH2:\(\sqrt{y}-1< 0\Leftrightarrow0\le y< 1\)
\(\Rightarrow D=\dfrac{-\left(\sqrt{y}-1\right)}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\dfrac{-1}{x-1}\)
Vậy...
\(E=\dfrac{1}{2x-1}.\sqrt{5x^4\left(1-4x+4x^2\right)}\)
\(=\dfrac{1}{2x-1}\sqrt{5x^4\left(2x-1\right)^2}=\dfrac{\sqrt{5}x^2\left|2x-1\right|}{2x-1}\)
TH1: \(2x-1>0\Leftrightarrow x>\dfrac{1}{2}\)
\(\Rightarrow E=\dfrac{\sqrt{5}x^2\left(2x-1\right)}{2x-1}=\sqrt{5}x^2\)
TH2:\(2x-1< 0\Leftrightarrow x< \dfrac{1}{2}\)
\(\Rightarrow E=\dfrac{-\sqrt{5}x^2\left(2x-1\right)}{2x-1}=-\sqrt{5}x^2\)
Vậy...
c)
\(C=\sqrt{\dfrac{x-6\sqrt{x}+9}{x+6\sqrt{x}+9}}=\sqrt{\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)^2}}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
d)
\(D=\dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}=\dfrac{x-1}{\sqrt{y}-1}.\dfrac{\left|\sqrt{y}-1\right|}{\left(x-1\right)^2}=\dfrac{\left|\sqrt{y}-1\right|}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
e)
\(E=\dfrac{1}{2x-1}.\sqrt{5x^4\left(1-4x+4x^2\right)}=\dfrac{1}{2x-1}.\sqrt{5x^4.\left(2x-1\right)^2}=\dfrac{1}{2x-1}.\sqrt{5}x^2.\left|2x-1\right|\)
