\(G=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\left(2x+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{4\sqrt{x}}\)
\(=\dfrac{2x+1}{4\sqrt{x}}\)
\(G=\dfrac{3}{4}\Rightarrow\dfrac{2x+1}{4\sqrt{x}}=\dfrac{3}{4}\Rightarrow2x+1=3\sqrt{x}\)
\(\Rightarrow2x-3\sqrt{x}+1=0\Rightarrow\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(G=\dfrac{2x+1}{4\sqrt{x}}\ge\dfrac{2\sqrt{2x}}{4\sqrt{x}}=\dfrac{\sqrt{2}}{2}\)
\(G_{min}=\dfrac{\sqrt{2}}{2}\) khi \(2x=1\Rightarrow x=\dfrac{1}{2}\)
a: Ta có: \(G=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{1-x}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)
\(=\dfrac{2x+1}{4\sqrt{x}}\)
b: Để \(G=\dfrac{3}{4}\) thì \(8x+4=12\sqrt{x}\)
\(\Leftrightarrow8x-12\sqrt{x}+4=0\)
\(\Leftrightarrow2x-3\sqrt{x}+1=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)=0\)
hay \(x=\dfrac{1}{4}\)
