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Sửa đề: \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}\)
a) Ta có: \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(=\dfrac{x-1}{x}\)
b) Sửa đề: \(2\sqrt{x+1}=5\)
Ta có: \(2\sqrt{x+1}=5\)
\(\Leftrightarrow\sqrt{x+1}=\dfrac{5}{2}\)
\(\Leftrightarrow x+1=\dfrac{25}{4}\)
hay \(x=\dfrac{21}{4}\)(thỏa ĐK)
Thay \(x=\dfrac{21}{4}\) vào biểu thức \(P=\dfrac{x-1}{x}\), ta được:
\(P=\left(\dfrac{21}{4}-1\right):\dfrac{21}{4}=\dfrac{17}{4}\cdot\dfrac{4}{21}=\dfrac{17}{21}\)
Vậy: Khi \(2\sqrt{x+1}=5\) thì \(P=\dfrac{17}{21}\)
c) Để \(P>\dfrac{1}{2}\) thì \(P-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2\left(x-1\right)}{2x}-\dfrac{x-1}{2x}>0\)
mà \(2x>0\forall x\) thỏa mãn ĐKXĐ
nen \(2\left(x-1\right)-x+1>0\)
\(\Leftrightarrow2x-2-x+1>0\)
\(\Leftrightarrow x-1>0\)
hay x>1
Kết hợp ĐKXĐ, ta được: x>1
Vậy: Để \(P>\dfrac{1}{2}\) thì x>1
a: ĐKXĐ: x>0; x<>4
\(P=\left(2-\sqrt{x}+2\right)\cdot\dfrac{1}{\sqrt{x}-2}=\dfrac{4-\sqrt{x}}{\sqrt{x}-2}\)
b: P=2/3
=>(4-căn x)/(căn x-2)=2/3
=>2căn x-4=12-3căn x
=>5căn x=16
=>x=256/25
c: Khi x=8-2căn 7 thì \(P=\dfrac{4-\sqrt{7}+1}{\sqrt{7}-1-2}=\dfrac{5-\sqrt{7}}{\sqrt{7}-3}=-4-\sqrt{7}\)
a: \(P=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Ta có: \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{1}{2\sqrt{x}}\right)\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)+8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{2\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)}{2\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{8x-8\sqrt{x}+8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-2-\sqrt{x}+2}\)
\(=\dfrac{16x-8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{2\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{2\left(16-8\sqrt{x}\right)}{\sqrt{x}+2}\)
\(=\dfrac{32-16\sqrt{x}}{\sqrt{x}+2}\)
\(a)C=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\dfrac{x-4}{\sqrt{4x}}\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}\right)\dfrac{x-4}{2\sqrt{x}}\\ =\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\right)\dfrac{x-4}{2\sqrt{x}}\\ =\dfrac{2x}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\\ =\dfrac{2x\left(x-4\right)}{2\sqrt{x}\left(x-4\right)}\\ =\sqrt{x}\)
b) C>3
\(\Rightarrow\sqrt{x}>3\\ \Leftrightarrow x>9\)
\(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}\left(x>0;x\ne1\right)\)
\(a,P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(=\dfrac{x-1}{x}\)
\(b,P>\dfrac{1}{2}\Rightarrow\dfrac{x-1}{x}>\dfrac{1}{2}\Leftrightarrow2x-2>x\Leftrightarrow x>2\)
\(a.P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{10\sqrt{x}-2x}\left(x>0,x\ne4,x\ne25\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}\right].\dfrac{x-4}{10\sqrt{x}-2x}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{10\sqrt{x}-2x}\)
\(=\dfrac{2x}{x-4}.\dfrac{x-4}{2\sqrt{x}\left(5-\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}}{5-\sqrt{x}}\)
\(b.\) Thay \(x=\dfrac{1}{4}\) vào P, ta được:
\(\dfrac{\sqrt{\dfrac{1}{4}}}{5-\sqrt{\dfrac{1}{4}}}=\dfrac{0,5}{5-0,5}=\dfrac{1}{9}\)
Vậy ......................
\(c.P< -1\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{5-\sqrt{x}}< -1\)
\(\Leftrightarrow\dfrac{\sqrt{x}+5-\sqrt{x}}{5-\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{5}{5-\sqrt{x}}< 0\)
\(\Leftrightarrow5-\sqrt{x}< 0\)
\(\Leftrightarrow\sqrt{x}>5\)
\(\Leftrightarrow x>25\left(tm\right)\)
Vậy ...................