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E = ( \(\dfrac{\sqrt{x}}{\sqrt{x-1}}\)- \(\dfrac{1}{x-\sqrt{x}}\)) : ( \(\dfrac{1}{\sqrt{x+1}}\)+\(\dfrac{2}{\sqrt{x-1}}\))
a) ta có ĐKXĐ của E là x \(\ne\) 1
x \(\ne\) 0
x \(\ne\) -1
b) ( \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)- \(\dfrac{\sqrt{1}}{x-\sqrt{x}}\)) : ( \(\dfrac{1}{\sqrt{x}+1}\)+\(\dfrac{2}{\sqrt{x}-1}\))
= (\(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)- \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) :(\(\dfrac{1}{\sqrt{x}+1}\)+ \(\dfrac{2}{x+1}\))
= ( \(\dfrac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) : (\(\dfrac{1\left(x-1\right)+2\sqrt{x+1}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\))
= ( \(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) : \(\dfrac{x-1+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
= \(\dfrac{1}{\sqrt{x}}\): \(\dfrac{\left(x-1\right)+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
= \(\dfrac{1}{\sqrt{x}}\). \(\dfrac{1}{2}\)
= \(\dfrac{1}{2\sqrt{x}}\)
a) Để biểu thức E được xác định thì \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\9x-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x>0\\x\ne\dfrac{1}{9}\end{matrix}\right.\)
b) \(E=\left(1-\dfrac{2\sqrt{x}}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{9x-1}\right):\left(\dfrac{9\sqrt{x}+6}{3\sqrt{x}+1}-3\right)=\left[\dfrac{3\sqrt{x}+1-2\sqrt{x}}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\dfrac{9\sqrt{x}+6-9\sqrt{x}-3}{3\sqrt{x}+1}\right)=\left[\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3}{3\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}.\left(1+\dfrac{1}{3\sqrt{x}-1}\right).\dfrac{3\sqrt{x}+1}{3}=\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}.\dfrac{3\sqrt{x}+1}{3}.\dfrac{3\sqrt{x}}{3\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
a. ĐKXĐ:x\(\ge\)3
b.ĐKXĐ:x\(\ge\)3
c.Mọi x đều thỏa mãn
d. ĐKXĐ: x>3
e.ĐKXĐ: x>3 hoặc x<-5
ĐKXĐ: \(x-\sqrt{\left(x-2\right)^2}>=0\)
=>x>=|x-2|
=>x^2>=(x-2)^2 và x>=0
=>0>=-4x+4 và x>=0
=>x>=0 và -4x+4<=0
=>x>=0 và -4x<=-4
=>x>=1
\(E=\sqrt{x-\sqrt{\left(x-2\right)^2}}=\sqrt{x-\left|x-2\right|}\)
\(DKXD:\left[{}\begin{matrix}x-x+2\ge0\\x+x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2\ge0\left(LD\right)\\2x\ge2\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)