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a) A=\(\left(\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\right)\times\frac{x-4}{\sqrt{x}+3}\) (dkxd:x # 4 ,x# 9)
=> A=\(\frac{\sqrt{x}+2-\sqrt{x}+2}{x-4}\times\frac{x-4}{\sqrt{x}+3}\)
=> A=\(\frac{4}{\sqrt{x}+3}\)
b) A>1/2 <=> \(\frac{4}{\sqrt{x}+3}>\frac{1}{2}\Leftrightarrow\sqrt{x}+3< 8\Leftrightarrow\sqrt{x}< 5\Leftrightarrow x< 25\) (tmdkxd)
Vay .....
a/ Ta có
P = \(\frac{1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) - \(\frac{2+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) - \(\frac{1+\sqrt{x}}{x+\sqrt{x}+1}\)
= \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}\)
1) Bạn đánh nhầm \(\sqrt{x}+3\rightarrow\sqrt{x+3}\); \(\sqrt{x}-3\rightarrow\sqrt{x-3}\)
Sửa : \(ĐKXĐ:x\ne\pm\sqrt{3}\)
a) \(M=\frac{x-\sqrt{x}}{x-9}+\frac{1}{\sqrt{x}+3}-\frac{1}{\sqrt{x}-3}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
b) Để \(M=\frac{3}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3}{4}\)
\(\Leftrightarrow4\sqrt{x}+8=3\sqrt{x}+9\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)(tm)
Vậy để \(A=\frac{3}{4}\Leftrightarrow x=1\)
c) Khi x = 4
\(\Leftrightarrow M=\frac{\sqrt{4}+2}{\sqrt{4}+3}\)
\(\Leftrightarrow M=\frac{2+2}{2+3}\)
\(\Leftrightarrow M=\frac{4}{5}\)
Vậy khi \(x=4\Leftrightarrow M=\frac{4}{5}\)
\(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right)\div\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Để A > 0
=> \(\frac{\sqrt{x}-1}{\sqrt{x}}>0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>1\\\sqrt{x}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x>0\end{cases}}\Leftrightarrow x>1\)
2. \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}< 0\end{cases}}\)( dễ thấy trường hợp này không xảy ra :> )
Vậy với x > 1 thì A > 0
ĐKXĐ: \(x\ge0;x\ne1\)
\(Q=\frac{\left(\sqrt{x}+1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{4\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-4\sqrt{x}-4}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-4}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-2}{x-1}\)
Để \(Q>1\Rightarrow\frac{-2}{x-1}>1\Rightarrow\frac{x+1}{x-1}< 0\Rightarrow x-1< 0\Rightarrow x< 1\)
Vậy \(0\le x< 1\)