a, \(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1-\dfrac{2\sqrt{x}}{x+1}\right)\)
\(=\left[\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right]:\dfrac{x+1-2\sqrt{x}}{x+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\dfrac{\left(\sqrt{x}-1\right)^2}{x+1}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
b, \(P\ge1\Leftrightarrow\dfrac{1}{\sqrt{x}-1}\ge1\)
\(\Leftrightarrow\dfrac{1-\sqrt{x}+1}{\sqrt{x}-1}\ge0\)
\(\Leftrightarrow\dfrac{2-\sqrt{x}}{\sqrt{x}-1}\ge0\)
\(\Leftrightarrow1< x\le4\)
a: Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1-\dfrac{2\sqrt{x}}{x+1}\right)\)
\(=\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\dfrac{x+1-2\sqrt{x}}{x+1}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
3.
a, \(A=\dfrac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(1-\dfrac{1}{x-1}\right)\)
\(=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}.\dfrac{x-1-1}{x-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}}{\sqrt{\left(x-2\right)^2}}.\dfrac{x-2}{x-1}\)
\(=\dfrac{\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|}{\left|x-2\right|}.\dfrac{x-2}{x-1}\)
Xem thử sai ở đâu không nha.
