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a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
a) ĐKXĐ : \(0\le x\ne4\)
b) \(A=\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{4\sqrt{x}-1}{x-4}\right):\frac{1}{x-4}\)
\(=\left[\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\left(x-4\right)\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(=\frac{-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=-1\)
\(A=\left[\frac{\left(\sqrt{x}-2\right)\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{x-4}\right]:\frac{1}{x-4}\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{x-4}.\left(x-4\right)\)=\(=\frac{-1}{x-4}.\left(x-4\right)=-1\)
Vậy giá trị của A thỏa mãn mọi x và rút gọn lại còn -1
A=\(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x-2}}\)
Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)
C=\(\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{1+x}{\sqrt{x}}\)
Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0
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 \(\left|P\right|=1\Leftrightarrow x\in\varnothing\)
Lời giải:
a)
\(A=\left[\frac{4\sqrt{x}(\sqrt{x}-2)}{(2+\sqrt{x})(\sqrt{x}-2)}-\frac{8x}{(\sqrt{x}-2)(\sqrt{x}+2)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-2)}-\frac{2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\right]\)
\(=\frac{4\sqrt{x}(\sqrt{x}-2)-8x}{(\sqrt{x}-2)(\sqrt{x}+2)}:\frac{\sqrt{x}-1-2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\)
\(=\frac{-4x-8\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}(\sqrt{x}-2)}{3-\sqrt{x}}\)
\(=\frac{-4\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}(\sqrt{x}-2)}{3-\sqrt{x}}=\frac{4x}{\sqrt{x}-3}\)
b)
Để $A=-1\Leftrightarrow \frac{4x}{\sqrt{x}-3}=-1$
$\Leftrightarrow 4x+\sqrt{x}-3=0$
$\Leftrightarrow (4\sqrt{x}-3)(\sqrt{x}+1)=0$
$\Rightarrow 4\sqrt{x}-3=0$
$\Leftrightarrow x=\frac{9}{16}$ (thỏa mãn ĐKXĐ)
Vậy........