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ĐKXĐ: \(x>0;x\ne1;x\ne9\)

\(B=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-1}\right)\)

\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{x-1-x+3}\)

\(=\frac{1}{\sqrt{x}}.\frac{\sqrt{x}-3}{2}\)

\(=\frac{\sqrt{x}-3}{2\sqrt{x}}\)

Để B < 0 thì 

\(\frac{\sqrt{x}-3}{2\sqrt{x}}< 0\) 

\(\Rightarrow\)\(\sqrt{x}-3\)và \(2\sqrt{x}\)trái dấu mà 

\(2\sqrt{x}\ge0\)\(\Rightarrow\sqrt{x}-3< 0\)

\(\Rightarrow\sqrt{x}< 3\)

\(\Rightarrow x< 9\)

23 tháng 9 2018

a) DK de P xác dinh : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

b) \(P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{1-x}+\frac{\left(\sqrt{x}-2\right)^2+3\sqrt{x}-x}{1-\sqrt{x}}\)

\(=\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{-\sqrt{x}+4}{1-\sqrt{x}}\)

\(=\frac{4}{1-\sqrt{x}}\)

c) de P > o thì \(1-\sqrt{x}>0\Rightarrow\sqrt{x}< 1\Rightarrow0< x< 1\)

15 tháng 7 2021

\(a,x>0;x\ne4,9\)

\(b,Q=\left(\frac{1}{\sqrt{x}-3}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\)

\(Q=\left(\frac{\sqrt{x}-\sqrt{x}+3}{x-3\sqrt{x}}\right):\left(\frac{x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(Q=\frac{3}{x-3\sqrt{x}}:\frac{-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(Q=\frac{3}{\sqrt{x}\left(\sqrt{x}-3\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{-5}\)

\(Q=\frac{3\sqrt{x}-6}{-5\sqrt{x}}\)

\(c,Q< 0< =>\frac{3\sqrt{x}-6}{-5\sqrt{x}}\)

\(-5\sqrt{x}< 0\)

\(< =>3\sqrt{x}-6>0\)

\(\sqrt{x}>2\)

\(x>4\)

P xác định khi \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)\)

\(=\frac{x-1}{\sqrt{x}}\)

P xác định khi \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)\)

\(=\frac{x-1}{\sqrt{x}}\)

21 tháng 10 2018

ĐKXĐ:   \(x\ge0;\)\(x\ne1\)

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

\(=\left(\frac{x}{\sqrt{x} \left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)

\(=\frac{x-1}{\sqrt{x}}\)

21 tháng 10 2018

a) bổ sung ĐKXĐ nhé:   \(x>0;\)\(x\ne1\)

b)  \(P< 0\)

=>  \(\frac{x-1}{\sqrt{x}}< 0\) 

=>  \(x-1< 0\)   (do \(\sqrt{x}>0\))

=>  \(x< 1\)

=>  \(0< x< 1\)