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a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
a: \(Q=\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(=\dfrac{a+2\sqrt{a}+1}{a-\sqrt{a}}\)
a: \(Q=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)
=\(\dfrac{\left(a-1\right)^2\cdot\left(-4\sqrt{a}\right)}{\left(a-1\right)\cdot4a}=\dfrac{-\left(a-1\right)}{\sqrt{a}}\)
b: Q<0
=>-(a-1)<0
=>a-1>0
=>a>1
c: Q=2
=>\(a-1=-2\sqrt{a}\)
=>\(a+2\sqrt{a}-1=0\)
=>\(\left[{}\begin{matrix}\sqrt{a}=-1+\sqrt{2}\left(nhận\right)\\\sqrt{a}=-1-\sqrt{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow a=3-2\sqrt{2}\)
a) ĐKXĐ: \(x>0;x\ne4\)
\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)
Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)
\(\text{#}\mathit{Toru}\)
Bài 1:
a) Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để Q dương thì \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
mà \(3\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
hay a>4
Kết hợp ĐKXĐ,ta được: a>4
Vậy: Để Q dương thì a>4
a) \(ĐK:x>0,x\ne1\)\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2}{x-1}\)
b) \(P=\dfrac{2}{x-1}\in Z\)
\(\Rightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp với đk
\(\Rightarrow x\in\left\{0;2;3\right\}\)
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) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
b) Ta có: \(Q-1=\dfrac{\sqrt{a}+1}{\sqrt{a}}-\dfrac{\sqrt{a}}{\sqrt{a}}=\dfrac{1}{\sqrt{a}}>0\forall a\) thỏa mãn ĐKXĐ
nên Q>1