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Sửa đề: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a+3\sqrt{a}+2-a+3\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{6\sqrt{a}}\)
\(=\dfrac{a-4}{6a\left(\sqrt{a}-1\right)}\)
c) Thay \(a=9-4\sqrt{5}\) vào Q, ta được:
\(Q=\dfrac{5-4\sqrt{5}}{6\left(9-4\sqrt{5}\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(9\sqrt{5}-27-20+12\sqrt{5}\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(21\sqrt{5}-47\right)}\)
\(=\dfrac{\left(5-4\sqrt{5}\right)\left(21\sqrt{5}+47\right)}{-24}\)
\(=\dfrac{105\sqrt{5}+235-420-188\sqrt{5}}{-24}\)
\(=\dfrac{-83\sqrt{5}-185}{-24}=\dfrac{83\sqrt{5}+185}{24}\)
a: Thay \(x=\dfrac{1}{4}\) vào A, ta được:
\(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)
b: Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)
\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-2}\)
c: Để B là số tự nhiên thì \(\sqrt{x}+4⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{1;2;3;6\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{3;4;5;8\right\}\)
hay \(x\in\left\{16;25;64\right\}\)
0 ; a ≠ 4 )
\(a,P=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}\\ P=\sqrt{a}+2+2+\sqrt{a}=2\sqrt{a}+4\\ b,P=a+1\Leftrightarrow a+1=2\sqrt{a}+4\\ \Leftrightarrow a-2\sqrt{a}-3=0\\ \Leftrightarrow\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)=0\\ \Leftrightarrow\sqrt{a}=3\left(\sqrt{a}\ge0\right)\\ \Leftrightarrow a=9\left(tm\right)\)
a) \(P=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{2-\sqrt{a}}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}\)
\(=\sqrt{a}+2+\sqrt{a}+2=2\sqrt{a}+4\)
b) \(P=a+1\Rightarrow2\sqrt{a}+4=a+1\Rightarrow a-2\sqrt{a}-3=0\)
\(\Rightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-3\right)=0\) mà \(\sqrt{a}+1>0\Rightarrow\sqrt{a}=3\Rightarrow a=9\)