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a: ĐKXĐ: x>0; x<>1
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\dfrac{1}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P<0 thì \(\sqrt{x}-1< 0\)
=>0<x<1
c: Để P là số nguyên thì \(\sqrt{x}-1+2⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2\right\}\)
hay \(x\in\left\{4;0;9\right\}\)
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne9\\x\ne4\end{cases}}\)
\(P=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{\sqrt{x}-3}{2\sqrt{x}-x}\)
\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(\Leftrightarrow P=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(\Leftrightarrow P=\frac{4x\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow P=\frac{4x}{\sqrt{x}-3}\)
b) Để P < 0
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-3< 0\Leftrightarrow4x>0\\\sqrt{x}-3>0\Leftrightarrow4x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< 3\Leftrightarrow x>0\\\sqrt{x}>3\Leftrightarrow x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x< 9\Leftrightarrow x>0\left(ktm\right)\\x>9\Leftrightarrow x< 0\left(ktm\right)\end{cases}}\)
Vậy để \(P< 0\Leftrightarrow x\in\varnothing\)
Để P > 0
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-3>0\Leftrightarrow4x>0\\\sqrt{x}-3< 0\Leftrightarrow4x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}>3\Leftrightarrow x>0\left(tm\right)\\\sqrt{x}< 3\Leftrightarrow x< 0\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow x>9\Leftrightarrow x>0\left(tm\right)\)
Vậy để \(P>0\Leftrightarrow x>9\)
c) Để \(\left|P\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}P=1\left(tm\right)\\P=-1\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow\frac{4x}{\sqrt{x}-3}=1\)
\(\Leftrightarrow4x=\sqrt{x}-3\)
\(\Leftrightarrow4x-\sqrt{x}+3=0\)
\(\Leftrightarrow\left(2\sqrt{x}-\frac{1}{4}\right)^2+\frac{47}{48}=0\left(ktm\right)\)
Vậy để \(\left|P\right|=1\Leftrightarrow x\in\varnothing\)
a/ \(A=\left(\dfrac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+3}{1-\sqrt{x}}\right)\cdot\left(\dfrac{x-1}{2x+\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right)\cdot\left(\dfrac{x-1}{2x+\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\cdot\dfrac{x-1}{2x+\sqrt{x}-1}=\dfrac{-3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(2x+2\sqrt{x}-\sqrt{x}-1\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}=\dfrac{-3}{2\sqrt{x}-1}\)
b/ \(A< 0\Leftrightarrow\dfrac{-3}{2\sqrt{x}-1}< 0\)
Ta thấy -3 < 0 nên để A < 0 thì:
\(2\sqrt{x}-1>0\)
\(\Leftrightarrow2\sqrt{x}>1\)
\(\Leftrightarrow\sqrt{x}>\dfrac{1}{2}\Leftrightarrow x>\dfrac{1}{4}\)
Vậy \(x>\dfrac{1}{4}\) thì A < 0
Bạn coi kĩ lại câu 1 đi bạn \(\sqrt{x}-2\) chứ không phải \(\sqrt{x-2}\)
Câu 1:
Với \(x>0,x\ne4\), ta có:
\(A=\left(\dfrac{x+2\sqrt{x}}{x-2\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-2}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{2}{\sqrt{x}-2}\)
b) Với \(x>0,x\ne4\): \(A< 0\)
\(\Rightarrow\dfrac{2}{\sqrt{x}-2}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\left(2>0\right)\)
\(\Leftrightarrow x< 4\)
Câu 2:
\(A=\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}\)
\(\Rightarrow\sqrt{2}A=\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}\)
\(\Rightarrow\sqrt{2}A=\sqrt{\left(\sqrt{17}+1\right)^2}-\sqrt{\left(\sqrt{17}-1\right)^2}\)
\(\Rightarrow\sqrt{2}A=\sqrt{17}+1-\sqrt{17}+1\)
\(\Rightarrow\sqrt{2}A=2\)
\(\Rightarrow A=\sqrt{2}\)
a) Để biểu thức P xác định thì \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Vậy ĐKXĐ:x\(\ge0\),x\(\ne9\)
\(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}=\left[\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{\left(-3\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)
b) Ta có \(P< \dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{1}{2}\Leftrightarrow-6< \sqrt{x}+3\Leftrightarrow\sqrt{x}>-9\)
Vì \(\sqrt{x}\ge0\) và 0>-9
Vậy \(x\ge0\)
Kết hợp với ĐKXĐ, Vậy \(x\ge0\) và \(x\ne9\) thì P<\(\dfrac{1}{2}\)
Để C<1 thì C-1<0
\(\Leftrightarrow\dfrac{-1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}-2}>0\)
=>x>4 hoặc x<1