Question

Rút gọn biểu thức:

A=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{x-1}\)với  x\(\ge\)0;\(x\ne\)1

Tìm x để A <\(\dfrac{3}{5}\)

Answer 1

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{x-1}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(A< \dfrac{3}{5}\Rightarrow\dfrac{3}{5}-A>0\Rightarrow\dfrac{3}{5}-\dfrac{\sqrt{x}-3}{\sqrt{x}-1}>0\)

\(\Rightarrow\dfrac{3\left(\sqrt{x}-1\right)-5\left(\sqrt{x}-3\right)}{5\left(\sqrt{x}-1\right)}>0\Rightarrow\dfrac{12-2\sqrt{x}}{5\left(\sqrt{x}-1\right)}>0\)

\(\Rightarrow\dfrac{2}{5}.\dfrac{6-\sqrt{x}}{\sqrt{x}-1}>0\Rightarrow\dfrac{6-\sqrt{x}}{\sqrt{x}-1}>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6-\sqrt{x}>0\\\sqrt{x}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}6-\sqrt{x}< 0\\\sqrt{x}-1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}1< x< 36\\\left\{{}\begin{matrix}x>36\\x< 1\end{matrix}\right.\left(l\right)\end{matrix}\right.\) 

\(\Rightarrow1< x< 36\)

 

Answer 2

\(=>A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

để \(A< \dfrac{3}{5}< =>\dfrac{\sqrt{x}-3}{\sqrt{x}-1}< \dfrac{3}{5}\)

\(< =>\dfrac{5\left(\sqrt{x}-3\right)-3\left(\sqrt{x}-1\right)}{5\left(\sqrt{x}-1\right)}< 0\)

\(< =>\dfrac{2\sqrt{x}-12}{5\left(\sqrt{x}-1\right)}< 0\)

\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}2\sqrt{x}-12>0\\5\left(\sqrt{x}-1\right)< 0\end{matrix}\right.\\\left[{}\begin{matrix}2\sqrt{x}-12< 0\\5\left(\sqrt{x}-1\right)>0\end{matrix}\right.\end{matrix}\right.\)\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}x>36\\x< 1\end{matrix}\right.\\\left[{}\begin{matrix}x< 36\\x>1\end{matrix}\right.\end{matrix}\right.=>1< x< 36\left(tm\right)\)

Answer 3

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{x-1}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)